RS Aggarwal Solutions for Class 12 Chapter 17 Area of Bounded Regions

 

Question 1. Find the area of the region bounded by the curve y = x², the x-axis, and the lines x = 1 and x = 3.
Answer: The region is enclosed by four boundaries - the curve y = x², the x-axis, and the vertical lines x = 1 and x = 3. When x = 3, the curve meets the line to form point A at (3, 9), and when x = 1, it forms point B at (1, 1). Points C and D lie on the x-axis at (1, 0) and (3, 0) respectively. The required area is calculated as the definite integral of x² from 1 to 3.

\[ \text{Area} = \int_1^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_1^3 = \left(\frac{27}{3} - \frac{1}{3}\right) = \frac{26}{3} \text{ sq. units} \]
In simple words: To find the area under a curve between two x-values, you integrate the function. The answer comes out to 26/3 square units, which is approximately 8.67 square units.

Exam Tip: Always set up the integral with the correct limits (lower and upper x-values) and remember to apply the power rule for integration: the integral of x^n is x^(n+1)/(n+1).

 

Question 2. Find the area of the region bounded by the parabola y² = 4x, the x-axis, and the lines x = 1 and x = 4.
Answer: This region is bounded by the parabola y² = 4x opening to the right, the x-axis, and two vertical lines. The parabola intersects x = 4 at point A where y = 4, and intersects x = 1 at point B where y = 2. Since the parabola is symmetric about the x-axis, the total area equals twice the area above the x-axis. By solving for y, we get y = √(4x) = 2√x. The area is computed by integrating this function from x = 1 to x = 4.

\[ \text{Area} = 2 \int_1^4 \sqrt{4x} \, dx = 2 \int_1^4 2\sqrt{x} \, dx = 4 \left[\frac{2x^{3/2}}{3}\right]_1^4 = 4 \left(\frac{16}{3} - \frac{2}{3}\right) = \frac{56}{3} \text{ sq. units} \]
In simple words: When a parabola is symmetric about the x-axis, calculate the area above the axis and double it. Convert y² = 4x into y = 2√x, then integrate from 1 to 4 to get about 18.67 square units.

Exam Tip: For curves of the form y² = 4ax, recognize the symmetry immediately and use the doubling technique to simplify calculations.

 

Question 3. Find the area under the curve y = √(6x + 4) (above the x-axis) from x = 0 to x = 2.
Answer: The curve y = √(6x + 4) forms a shape bounded by the axes and the vertical line x = 2. Rearranging gives y² = 6x + 4, which represents a parabola opening rightward with vertex at (0, -2/3). However, only the upper half above the x-axis is considered. The four boundary points are O at the origin, A where the curve meets x = 0, B where the curve meets x = 2, and C where x = 2 meets the x-axis. The area is found by integrating the function from 0 to 2.

\[ \text{Area} = \int_0^2 \sqrt{6x + 4} \, dx = \left[\frac{(6x + 4)^{3/2}}{9}\right]_0^2 = \frac{1}{9}\left[(16)^{3/2} - (4)^{3/2}\right] = \frac{1}{9}(64 - 8) = \frac{56}{9} = \frac{28}{3} \text{ sq. units} \]
In simple words: To integrate a square root function with a linear expression inside, use substitution or apply the chain rule in reverse. The result is approximately 9.33 square units.

Exam Tip: When integrating √(ax + b), raise the expression to the power 3/2 and divide by 3a/2, then apply the limits carefully.

 

Question 4. Determine the area enclosed by curve y = x³, and the lines y = 0, x = 2 and x = 4.
Answer: The region is bounded by the cubic curve y = x³, the x-axis (y = 0), and two vertical boundaries at x = 2 and x = 4. Point A is where the curve meets x = 2, giving y = 8. Point B is where the curve meets x = 4, giving y = 64. Points C and D are on the x-axis at (4, 0) and (2, 0) respectively. The area is the definite integral of x³ from 2 to 4.

\[ \text{Area} = \int_2^4 x^3 \, dx = \left[\frac{x^4}{4}\right]_2^4 = \frac{1}{4}(256 - 16) = \frac{240}{4} = 60 \text{ sq. units} \]
In simple words: For a cubic function, raise the power to 4 and divide by 4. Then subtract the lower limit result from the upper limit result to get the area of 60 square units.

Exam Tip: Always evaluate the antiderivative at both limits and subtract; common errors include forgetting to subtract or making arithmetic mistakes in the powers.

 

Question 5. Determine the area under the curve y = √(a² - x²), included between the lines x = 0 and x = 4.
Answer: The equation y = √(a² - x²) can be rewritten as x² + y² = a², which represents a circle with center at the origin and radius a units. Since the problem specifies x from 0 to 4, and both x and y have even powers, the curve is symmetric about both axes. The region extends from the y-axis (x = 0) to the line x = 4. The three boundary points are O at the origin, B where x = 4 meets the circle, and C where x = 0 meets the curve. The area is computed using the standard integral formula for circular regions.

\[ \text{Area} = \int_0^4 \sqrt{a^2 - x^2} \, dx = \left[\frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\right]_0^4 = \frac{a^2\pi}{4} \text{ sq. units} \]
In simple words: For a semicircular region, use the inverse sine formula. The area equals (a²/2) times [90° or π/2 radians] minus zero, giving (a²π)/4 square units.

Exam Tip: Recognize circle equations immediately and apply the inverse trigonometric formula; remember that sin⁻¹(1) = 90° and sin⁻¹(0) = 0°.

 

Question 6. Using integration, find the area of the region bounded by the lines 2y = 5x + 7, the x-axis and the lines x = 2 and x = 8.
Answer: The region is enclosed by a straight line, the x-axis, and two vertical boundaries. The line 2y = 5x + 7 can be rewritten as y = (5x + 7)/2. When x = 2, y = 17/2 = 8.5 (point A). When x = 8, y = 47/2 = 23.5 (point B). Points C and D are on the x-axis at (8, 0) and (2, 0) respectively. This forms a trapezoid. The area is found by integrating the line equation from x = 2 to x = 8.

\[ \text{Area} = \int_2^8 \frac{5x + 7}{2} \, dx = \frac{1}{2}\left[\frac{5x^2}{2} + 7x\right]_2^8 = \frac{1}{2}\left[\left(160 + 56\right) - \left(10 + 14\right)\right] = \frac{1}{2}(192) = 96 \text{ sq. units} \]
In simple words: For a linear boundary, set up the integral of the line equation. Integrate to get a quadratic expression, then evaluate at both limits and find the difference to obtain the area of 96 square units.

Exam Tip: For trapezoid-shaped regions formed by a line and the x-axis, integration works perfectly - the result equals (1/2) × (sum of parallel sides) × height only if it's a simple shape.

 

Question 7. Find the area of the region bounded by the curve y² = 4x and the lines x = 3.
Answer: The parabola y² = 4x opens to the right from the origin. The vertical line x = 3 intersects the parabola at two points due to symmetry about the x-axis. Point A is at (3, 2√3) where y is positive, and point B is at (3, -2√3) where y is negative. Point C is on the x-axis at (3, 0), and O is the origin. Because the parabola is symmetric about the x-axis, the total area equals twice the area of the upper half. By solving y² = 4x for y, we get y = 2√x. The area calculation involves integrating from 0 to 3.

\[ \text{Area} = 2 \int_0^3 2\sqrt{x} \, dx = 4\left[\frac{2x^{3/2}}{3}\right]_0^3 = \frac{8}{3}(3\sqrt{3}) = 8\sqrt{3} \text{ sq. units} \]
In simple words: Since the parabola is symmetric about the x-axis, find the area above the axis and multiply by 2. Convert y² = 4x to y = 2√x, integrate from 0 to 3, and double the result to get approximately 13.86 square units.

Exam Tip: Always check for symmetry first - it halves your work by allowing you to calculate one half and multiply by 2.

 

Question 8. Evaluate the area bounded by the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) above the x-axis.
Answer: The ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) has its center at the origin with a = 2 (semi-minor axis along x) and b = 3 (semi-major axis along y). The ellipse intersects the x-axis at A(-2, 0) and B(2, 0), and intersects the y-axis at C(0, 3) and D(0, -3). The region above the x-axis forms the upper half of the ellipse. Due to symmetry about the y-axis, the total area equals twice the area from x = 0 to x = 2. Solving for y gives \( y = \frac{3}{2}\sqrt{4 - x^2} \).

\[ \text{Area} = 2 \int_0^2 \frac{3}{2}\sqrt{4 - x^2} \, dx = 3\left[\frac{x\sqrt{4 - x^2}}{2} + 2\sin^{-1}\left(\frac{x}{2}\right)\right]_0^2 = 3\left[\pi\right] = 3\pi \text{ sq. units} \]
In simple words: For an ellipse, extract the semi-axes from the equation. The upper half has area equal to half the total ellipse area, which is (π × a × b)/2 = 3π square units.

Exam Tip: The area of an ellipse is πab; the area of the upper half is (πab)/2. Always identify a and b correctly from the standard form.

 

Question 9. Using integration, find the area of the region bounded by the lines y = 1 + |x + 1|, x = -2, x = 3 and y = 0.
Answer: The absolute value function y = 1 + |x + 1| changes form at x = -1. For x ≥ -1, it becomes y = x + 2, and for x < -1, it becomes y = -x. The region must be split into two parts: from x = -2 to x = -1 (below the line y = -x) and from x = -1 to x = 3 (below the line y = x + 2). At x = -2, y = 2 (point C). At x = 3, y = 5 (point B). Point F is at (-1, 0) where the V-shaped boundary touches the x-axis. Points A and D are at (3, 0) and (-2, 0) respectively. The total area is found by integrating each piece separately and adding them.

\[ \text{Area} = \int_{-2}^{-1} (-x) \, dx + \int_{-1}^{3} (x + 2) \, dx = \left[-\frac{x^2}{2}\right]_{-2}^{-1} + \left[\frac{x^2}{2} + 2x\right]_{-1}^{3} = \frac{3}{2} + \frac{27}{2} = \frac{27}{2} \text{ sq. units} \]
In simple words: When dealing with absolute value functions, split the integration at the point where the expression inside becomes zero. Integrate each part separately using its specific formula, then add the results together.

Exam Tip: Always identify where the absolute value expression equals zero - this is your splitting point. Set up two separate integrals with different limits and functions.

 

Question 10. Find the area bounded by the curve y = (4 - x²), the y-axis and the lines y = 0, y = 3.
Answer: The curve y = 4 - x² is a downward-opening parabola with vertex at (0, 4). Rearranging gives x² = 4 - y, or \( x = \sqrt{4 - y} \). The curve is symmetric about the y-axis since x has an even power. Integration proceeds with respect to y from 0 to 3. The boundaries are: point A at (-√1, 0) where the curve meets y = 0, point B at (-1, 3) where the curve meets y = 3, point C at (1, 3) where the curve meets y = 3 on the positive side, and point D at (√1, 0). Since the region is symmetric about the y-axis, the total area equals twice the area from x = 0 to x = √(4 - y).

\[ \text{Area} = \int_0^3 2\sqrt{4 - y} \, dy = 2\left[-\frac{2}{3}(4 - y)^{3/2}\right]_0^3 = -\frac{4}{3}[(1) - (8)] = \frac{28}{3} \text{ sq. units} \]
In simple words: When a parabola opens sideways and you need area between two y-values, integrate with respect to y instead of x. Use the symmetry about the y-axis to simplify the calculation.

Exam Tip: Always match the variable of integration to the setup - if limits are given as y-values, integrate dy; if as x-values, integrate dx.

 

Question 11. Using integration, find the area of the region bounded by the triangle whose vertices are A(-1, 2), B(1, 6) and C(3, 4).
Answer: A triangle with vertices A(-1, 2), B(1, 6), and C(3, 4) can be analyzed by examining the three sides separately. The side AB has the equation y = (3x + 7)/2, derived from the two-point form. The side BC has the equation y = (11 - x)/2. The side CA has the equation y = (x + 5)/2. The region between the triangle and the x-axis is partitioned into three areas: ABDE (under line AB), BCFD (under line BC), and CAFE (under line CA). By integrating each line over its respective x-interval and using the standard power rule, the areas are: Area under AB = 7 sq. units, Area under BC = 9 sq. units, and Area under CA = (calculated separately). The final triangle area equals the sum of these minus overlapping regions, yielding the net enclosed area.

\[ \text{Area of triangle} = \text{Area under AB} + \text{Area under BC} - \text{Area under CA} = 7 + 9 - (7 + 9 - \text{area}) \]

Using the exact integral limits and vertex coordinates:
\[ \text{Total Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2}|(-1)(2) + 1(2) + 3(-4)| = 8 \text{ sq. units} \]
In simple words: Find equations of the three sides using the two-point form, integrate each line over its x-range, then use the relationship between the areas to find the triangle's enclosed area. Alternatively, use the determinant formula with vertices.

Exam Tip: For triangle areas using integration, set up each side's equation carefully and track which regions are above and below the x-axis to avoid sign errors.

 

Question 12. Using integration, find the area of ΔABC, the equation of whose sides AB, BC and AC are given by y = 4x + 5, x + y = 5 and 4y = x + 5 respectively.
Answer: We are given:

• Triangle ABC
• Side AB has equation y = 4x + 5
• Side BC has equation x + y = 5
• Side CA has equation 4y = x + 5

Finding vertex B by solving AB and BC:
From AB: y = 4x + 5 and BC: y = 5 - x
4x + 5 = 5 - x
5x = 0, so x = 0
Substituting x = 0 into AB gives y = 5
Thus B = (0, 5)

Finding vertex C by solving BC and CA:
From BC: y = 5 - x and CA: 4y = x + 5
4(5 - x) = x + 5
20 - 4x = x + 5
15 = 5x, so x = 3
Substituting x = 3 into BC gives y = 2
Thus C = (3, 2)

Finding vertex A by solving AB and CA:
From AB: y = 4x + 5 and CA: 4y = x + 5
4(4x + 5) = x + 5
16x + 20 = x + 5
15x = -15, so x = -1
Substituting x = -1 into AB gives y = 1
Thus A = (-1, 1)

The area under line AB from x = -1 to x = 0:
\[ \text{Area under AB} = \int_{-1}^{0} (4x + 5) \, dx = \left[ 2x^2 + 5x \right]_{-1}^{0} = 0 - (2 - 5) = 3 \text{ sq. units} \quad \text{...(1)} \]

The area under line BC from x = 0 to x = 3:
\[ \text{Area under BC} = \int_{0}^{3} (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_{0}^{3} = 15 - \frac{9}{2} = \frac{21}{2} \text{ sq. units} \quad \text{...(2)} \]

The area under line CA from x = -1 to x = 3:
\[ \text{Area under CA} = \int_{-1}^{3} \frac{x + 5}{4} \, dx = \frac{1}{4} \left[ \frac{x^2}{2} + 5x \right]_{-1}^{3} = \frac{1}{4} \left[ \left( \frac{9}{2} + 15 \right) - \left( \frac{1}{2} - 5 \right) \right] = \frac{1}{4} \cdot 24 = 6 \text{ sq. units} \quad \text{...(3)} \]

Since the area under CA overlaps with the areas under AB and BC, we find the triangle's area by:
\[ \text{Area of } \triangle ABC = \text{Area under AB} + \text{Area under BC} - \text{Area under CA} \]
\[ = 3 + \frac{21}{2} - 6 = \frac{6 + 21 - 12}{2} = \frac{15}{2} \text{ sq. units} \]
In simple words: Find where the three lines meet to get the vertices. Then use integration to calculate the area beneath each line between its endpoints. Subtract the overlapping area to get the triangle's final area.

Exam Tip: Always verify your intersection points by substituting back into both line equations. This prevents errors that cascade through the entire integration.

 

Question 13. Using integration, find the area of the region bounded between the line x = 2 and the parabola y² = 8x.
Answer: We are given:

• The parabola y² = 8x
• The line x = 2 (parallel to the y-axis)

Properties of the parabola y² = 8x:

• Vertex is at the origin (0, 0)
• Since y has an even power, the parabola is symmetric about the x-axis

Key points of intersection:

• Point A: where y² = 8x and x = 2 meet with positive y: y² = 16, so y = 4. Thus A = (2, 4)
• Point B: where y² = 8x and x = 2 meet with negative y: y = -4. Thus B = (2, -4)
• Point C: where x = 2 meets the x-axis: C = (2, 0)

Since the region is symmetric about the x-axis:
\[ \text{Area of region OACB} = 2 \times \text{Area of region OAC} \]
\[ = 2 \int_{0}^{2} \sqrt{8x} \, dx = 2 \int_{0}^{2} 2\sqrt{2x} \, dx \]
\[ = 4\sqrt{2} \int_{0}^{2} \sqrt{x} \, dx = 4\sqrt{2} \left[ \frac{2x^{3/2}}{3} \right]_{0}^{2} \]
\[ = 4\sqrt{2} \cdot \frac{2 \cdot 2^{3/2}}{3} = \frac{8\sqrt{2}}{3} \left( 2\sqrt{2} \right) = \frac{32}{3} \text{ sq. units} \]
In simple words: The parabola opens to the right. Use integration to find the area on one side of the x-axis, then double it because of symmetry.

Exam Tip: Recognize symmetry early - it simplifies calculations by allowing you to integrate only half the region and multiply by 2.

 

Question 14. Using integration, find the area of region bounded by the line y - 1 = x, the x-axis, and the ordinates x = -2 and x = 3.
Answer: We are given:

• The line equation y = x + 1
• The x-axis (y = 0)
• The ordinate x = -2 (line parallel to y-axis)
• The ordinate x = 3 (line parallel to y-axis)

The four vertices of the region are:

• Point A: where y = x + 1 and x = 3 meet: A = (3, 4)
• Point B: where y = x + 1 and x = -2 meet: B = (-2, -1)
• Point C: where the x-axis and x = -2 meet: C = (-2, 0)
• Point D: where the x-axis and x = 3 meet: D = (3, 0)

Since the line crosses the x-axis when x = -1, the region splits into two parts:
\[ \text{Area of ABCD} = \int_{-2}^{-1} [-(x + 1)] \, dx + \int_{-1}^{3} (x + 1) \, dx \]
\[ = \int_{-2}^{-1} (-x - 1) \, dx + \int_{-1}^{3} (x + 1) \, dx \]
\[ = \left[ -\frac{x^2}{2} - x \right]_{-2}^{-1} + \left[ \frac{x^2}{2} + x \right]_{-1}^{3} \]
\[ = \left[ (-\frac{1}{2} + 1) - (-2 + 2) \right] + \left[ (\frac{9}{2} + 3) - (\frac{1}{2} - 1) \right] \]
\[ = \frac{1}{2} + \frac{15}{2} = 8.5 \text{ sq. units} \]
In simple words: The line enters the region below the x-axis (from x = -2 to x = -1) and above it (from x = -1 to x = 3). Calculate each part separately and add them together.

Exam Tip: Identify where the curve crosses the boundary line (x-axis in this case) and split the integral at that point to handle sign changes correctly.

 

Question 15. Sketch the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 2 and y = 4. Find the area of the region using integration.
Answer: We are given:

• The curve y = 4x²
• x = 0 (the y-axis)
• y = 2 (line parallel to x-axis)
• y = 4 (line parallel to x-axis)
• Only the region in the 1st quadrant is required

Properties of the curve y = 4x²:

• Since x has an even power, the curve is symmetric about the y-axis

Boundary points of the region:

• Point A: where y = 4x² and the y-axis meet at y = 4: A = (0, 4)
• Point B: where y = 4x² and y = 4 meet: 4x² = 4, so x = 1. Thus B = (1, 4)
• Point C: where y = 4x² and y = 2 meet: 4x² = 2, so x = 1/√2. Thus C ≈ (0.707, 2)
• Point D: where the y-axis and y = 2 meet: D = (0, 2)

We integrate with respect to y since the region is bounded by horizontal lines:
\[ x = \frac{1}{2}\sqrt{y} \]
\[ \text{Area of ABCD} = \int_{2}^{4} \frac{1}{2}\sqrt{y} \, dy = \frac{1}{2} \left[ \frac{2y^{3/2}}{3} \right]_{2}^{4} \]
\[ = \frac{1}{3} \left[ 4^{3/2} - 2^{3/2} \right] = \frac{1}{3} \left[ 8 - 2\sqrt{2} \right] \text{ sq. units} \]
In simple words: When the region is bounded by horizontal lines, it is easier to integrate with respect to y rather than x. Solve for x from the parabola equation and integrate between the y-values.

Exam Tip: Choose your integration variable based on the region's boundaries - if horizontal lines dominate, integrate with respect to y; if vertical lines dominate, integrate with respect to x.

 

Question 16. Sketch the region lying in the first quadrant and bounded by y = 9x², x = 0, y = 1 and y = 4. Find the area of the region, using integration.
Answer: We are given:

• The curve y = 9x²
• x = 0 (the y-axis)
• y = 1 (line parallel to x-axis)
• y = 4 (line parallel to x-axis)
• Only the region in the 1st quadrant is required

Properties of the curve y = 9x²:

• Since x has an even power, the curve is symmetric about the y-axis

Boundary points of the region:

• Point A: where y = 9x² and the y-axis meet at y = 4: A = (0, 4)
• Point B: where y = 9x² and y = 4 meet: 9x² = 4, so x = 2/3. Thus B = (2/3, 4)
• Point C: where y = 9x² and y = 1 meet: 9x² = 1, so x = 1/3. Thus C = (1/3, 1)
• Point D: where the y-axis and y = 1 meet: D = (0, 1)

We integrate with respect to y since the region is bounded by horizontal lines:
From y = 9x², we get \( x = \frac{1}{3}\sqrt{y} \)
\[ \text{Area of ABCD} = \int_{1}^{4} \frac{1}{3}\sqrt{y} \, dy = \frac{1}{3} \left[ \frac{2y^{3/2}}{3} \right]_{1}^{4} \]
\[ = \frac{2}{9} \left[ 4^{3/2} - 1^{3/2} \right] = \frac{2}{9} \left[ 8 - 1 \right] = \frac{14}{9} \text{ sq. units} \]
In simple words: Extract x from the parabola equation in terms of y. Then integrate with respect to y between the given limits to find the area between the horizontal lines.

Exam Tip: For parabolas of the form y = ax², convert to \( x = \frac{1}{\sqrt{a}}\sqrt{y} \) and integrate with respect to y when horizontal bounds are given - this avoids messy algebra.

 

Question 17. Find the area of the region enclosed between the circles x² + y² = 1 and (x - 1)² + y² = 1.
Answer: We are given:

• First circle: x² + y² = 1
• Second circle: (x - 1)² + y² = 1

Properties of the circles:

• First circle: center at (0, 0), radius = 1 unit
• Second circle: center at (1, 0), radius = 1 unit

Finding points of intersection:
Substitute y² = 1 - x² from the first circle into the second:
(x - 1)² + (1 - x²) = 1
x² - 2x + 1 + 1 - x² = 1
-2x + 2 = 1
\( x = \frac{1}{2} \)

Substitute x = 1/2 into the first circle:
\( y = \pm\frac{\sqrt{3}}{2} \)

Points of intersection: \( A = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) and \( B = \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \)

Line AB intersects the x-axis at \( D = \left( \frac{1}{2}, 0 \right) \)

Since both circles are symmetric about both axes, we can use:
\[ \text{Area of enclosed region} = 4 \times \text{Area of ADC} \]
where C is the intersection of the first circle with the positive x-axis at (1, 0).

From the first circle: \( y = \sqrt{1 - x^2} \)
\[ \text{Area of ADC} = \int_{1/2}^{1} \sqrt{1 - x^2} \, dx \]
\[ = \left[ \frac{x\sqrt{1 - x^2}}{2} + \frac{1}{2}\sin^{-1}(x) \right]_{1/2}^{1} \]
\[ = \left[ 0 + \frac{\pi}{4} \right] - \left[ \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}\left(\frac{1}{2}\right) \right] \]
\[ = \frac{\pi}{4} - \left[ \frac{\sqrt{3}}{4} + \frac{\pi}{12} \right] = \frac{\pi}{6} - \frac{\sqrt{3}}{4} \]
\[ \text{Area of enclosed region} = 4 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = \frac{2\pi}{3} - \sqrt{3} \text{ sq. units} \]
In simple words: Find where the circles intersect. Use symmetry to calculate only one section, then multiply by 4 to get the total area. The enclosed region is a lens shape where both circles overlap.

Exam Tip: Always exploit symmetry in circle problems - it cuts your work in half or even quarters. Mark key intersection points and axis intercepts clearly on a sketch.

 

Question 18. Sketch the region common to the circle x² + y² = 16 and the parabola x² = 6y. Also, find the area of the region, using integration.
Answer: We are given:

• The circle: x² + y² = 16
• The parabola: x² = 6y

Properties:

• Circle: center at (0, 0), radius = 4 units
• Parabola: vertex at (0, 0), opens upward, symmetric about the y-axis

Finding points of intersection:
Substitute x² = 6y into the circle equation:
6y + y² = 16
y² + 6y - 16 = 0
(y - 2)(y + 8) = 0
y = 2 or y = -8

Since y must be non-negative for the parabola (in the given region), y = 2
From x² = 6y: x² = 12, so x = ±2√3

Points of intersection: A = (2√3, 2) and B = (-2√3, 2)

From the circle: \( y = \sqrt{16 - x^2} \)
From the parabola: \( y = \frac{x^2}{6} \)

The perpendicular from A to the x-axis meets it at D = (2√3, 0)

Using symmetry about the y-axis:
\[ \text{Area of region} = 2 \times \text{Area of OCAO} \]
where C is at (4, 0) - the right intersection of the circle with the x-axis.

\[ \text{Area of OCAO} = \int_{0}^{2\sqrt{3}} \sqrt{16 - x^2} \, dx - \int_{0}^{2\sqrt{3}} \frac{x^2}{6} \, dx \]
\[ = \left[ \frac{x\sqrt{16 - x^2}}{2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{2\sqrt{3}} - \left[ \frac{x^3}{18} \right]_{0}^{2\sqrt{3}} \]
\[ = \left[ \sqrt{3} \cdot 2 + 8\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \right] - \frac{8\sqrt{3} \cdot 3}{18} \]
\[ = 2\sqrt{3} + 8 \cdot \frac{\pi}{3} - \frac{4\sqrt{3}}{3} = \frac{2\sqrt{3}}{3} + \frac{8\pi}{3} \]
\[ \text{Area of region} = 2 \left( \frac{2\sqrt{3}}{3} + \frac{8\pi}{3} \right) = \frac{4\sqrt{3} + 16\pi}{3} \text{ sq. units} \]
In simple words: Find where the circle and parabola intersect by substitution. Then integrate the difference between the circle and parabola equations over the appropriate interval. Use symmetry to simplify.

Exam Tip: When finding the area between two curves, always integrate (upper curve - lower curve). Identify which is which in your region of interest by testing a point.

 

Question 19. Sketch the region common to the circle x² + y² = 25 and the parabola y² = 8x. Also, find the area of the region, using integration.
Answer: We are given:

• The circle: x² + y² = 25
• The parabola: y² = 8x

Properties:

• Circle: center at (0, 0), radius = 5 units
• Parabola: vertex at (0, 0), opens to the right, symmetric about the x-axis

Finding points of intersection:
Substitute y² = 8x into the circle equation:
x² + 8x = 25
x² + 8x - 25 = 0
Using the quadratic formula:
\( x = \frac{-8 + \sqrt{64 + 100}}{2} = \frac{-8 + 2\sqrt{41}}{2} = -4 + \sqrt{41} \)
(The negative root is rejected since we need x ≥ 0 for the parabola region)

The circle intersects the x-axis at C = (5, 0) and E = (-5, 0)

From the circle: \( y = \sqrt{25 - x^2} \)
From the parabola: \( y = \sqrt{8x} \)

Let a = √41 - 4 (the x-coordinate of the intersection point)

Using symmetry about the x-axis:
\[ \text{Area of region} = 2 \times \text{Area of OCAO} \]
\[ \text{Area of OCAO} = \int_{0}^{a} \sqrt{8x} \, dx + \int_{a}^{5} \sqrt{25 - x^2} \, dx \]
\[ = 2\sqrt{2} \left[ \frac{2x^{3/2}}{3} \right]_{0}^{a} + \left[ \frac{x\sqrt{25 - x^2}}{2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{a}^{5} \]

After substitution and simplification:
\[ \text{Area of region} = \frac{16\pi + 4\sqrt{3}}{3} \text{ sq. units} \]
In simple words: Substitute the parabola into the circle equation to find intersection points. Integrate the parabola from the origin to the intersection, then integrate the circle from the intersection to its x-intercept. Double the result due to symmetry about the x-axis.

Exam Tip: For parabolas opening horizontally, integrate with respect to y when possible, or split the integration into two parts with different functions if the curve changes.

 

Question 20. Draw a rough sketch of the region and find the area enclosed by the region, using the method of integration.
Answer: We are given that the region's boundaries are \( R = \{(x,y): y^2 \leq 3x, 3x^2 + 3y^2 \leq 16\} \). This can be split into two parts: \( R_1 = \{(x,y): y^2 \leq 3x\} \) and \( R_2 = \{(x,y): 3x^2 + 3y^2 \leq 16\} \). From \( R_1 \), the equation \( y^2 = 3x \) represents a parabola with its vertex at the origin and symmetric about the x-axis due to the even power of y. From \( R_2 \), the equation \( 3x^2 + 3y^2 = 16 \) represents a circle with its vertex at the origin and a radius of \( \frac{4}{\sqrt{3}} \) units. To locate the intersection points of these curves, we substitute \( y^2 = 3x \) into the circle equation: \( 3x^2 + 9x - 16 = 0 \), which gives \( x = \frac{-9+\sqrt{273}}{6} \) (taking the positive value). When \( y = 0 \), the circle intersects the x-axis at \( C = (\frac{4}{\sqrt{3}}, 0) \) and \( E = (-\frac{4}{\sqrt{3}}, 0) \). The area we seek is the region bounded by both curves. We can write \( y = \sqrt{3x} \) for the parabola and \( y = \sqrt{\frac{16}{3} - x^2} \) for the circle. The total area of the required region is \( \frac{8\sqrt{2}}{3}\left(a^{\frac{3}{2}}\right) - \left[\frac{25\pi}{4}\right] + \left[a\sqrt{\frac{16}{3} - (a)^2} + \frac{25}{3}\sin^{-1}\left(\frac{\sqrt{3}a}{4}\right)\right] \) square units, where \( a = \frac{-9+\sqrt{273}}{6} \).
In simple words: The region is enclosed by a parabola that opens to the right and a circle. To find the area, we first find where they meet, then integrate. The parabola portion contributes one piece of area, and the circular portion contributes another. Adding them together gives the total enclosed area.

Exam Tip: Always identify the curves clearly and find their intersection points first - this tells you the bounds for integration. Remember to handle symmetric regions carefully by doubling the area of one half if needed.

 

Question 21. Draw a rough sketch and find the area of the region bounded by the parabolas y² = 4x and x² = 4y, using the method of integration.
Answer: We are given the boundaries: the first parabola is \( y^2 = 4x \) and the second parabola is \( x^2 = 4y \). The first parabola has its vertex at the origin and is symmetric about the x-axis since y has an even power. The second parabola also has its vertex at the origin and is symmetric about the y-axis since x has an even power. To find where they intersect, we substitute \( y = \frac{x^2}{4} \) from the second equation into the first: \( \left(\frac{x^2}{4}\right)^2 = 4x \), which simplifies to \( x^4 = 64x \). This gives us \( x(x^3 - 64) = 0 \), so \( x = 0 \) or \( x = 4 \). When \( x = 4 \), we get \( y = 4 \), so the two intersection points are \( A = (4,4) \) and \( O = (0,0) \). We can rewrite the parabolas as \( y = 2\sqrt{x} \) and \( y = \frac{x^2}{4} \). The region bounded by these curves has area equal to the integral \( \int_0^4 2\sqrt{x} \, dx - \frac{1}{4}\int_0^4 x^2 \, dx \). Computing the first integral: \( 2 \cdot \frac{2}{3}x^{3/2}\big|_0^4 = \frac{4}{3}(8) = \frac{32}{3} \). Computing the second integral: \( \frac{1}{4} \cdot \frac{x^3}{3}\big|_0^4 = \frac{1}{12}(64) = \frac{16}{3} \). Therefore, the required area is \( \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \) square units.
In simple words: Two parabolas meet at the origin and at the point (4,4). One opens rightward and the other opens upward. To find the area between them, we integrate the difference of the two functions from x = 0 to x = 4. The result gives us the trapped region's area.

Exam Tip: When two curves intersect at multiple points, carefully determine which curve is above (or to the right of) the other in the region of interest. Set up the integral with the outer curve minus the inner curve.

 

Question 22. Find by integration the area bounded by the curve y² = 4ax and the lines y = 2a and x = 0.
Answer: The given curve is \( y^2 = 4ax \), a parabola opening to the right. Since y appears with an even power, the curve distributes equally on both sides of the x-axis. The line \( y = 2a \) is parallel to the x-axis at a distance of 2a units from it. The line \( x = 0 \) is the y-axis. The curve \( y^2 = 4ax \) and the line \( y = 2a \) intersect when \( (2a)^2 = 4ax \), giving us \( x = a \). So point A is at \( (a, 2a) \). The curve meets the y-axis (where x = 0) at point B, which is at \( (0, 2a) \). The origin O is at \( (0, 0) \). From the parabola equation, we can write \( x = \frac{y^2}{4a} \). The region OBA is bounded by the y-axis (from O to B), the horizontal line (from B to A), and the parabolic curve (from A back to O). The area of region OBA is \( \int_0^{2a} x \, dy = \int_0^{2a} \frac{y^2}{4a} \, dy = \frac{1}{4a} \cdot \frac{y^3}{3}\big|_0^{2a} = \frac{1}{4a} \cdot \frac{(2a)^3}{3} = \frac{1}{4a} \cdot \frac{8a^3}{3} = \frac{2a^2}{3} \) square units.
In simple words: A parabola opens to the right from the y-axis. A horizontal line cuts through it. The bounded region is between the y-axis, the parabola, and the horizontal line. Integrating with respect to y (since the boundary is horizontal) makes the calculation straightforward.

Exam Tip: When a region is bounded by a horizontal or vertical line, choose the integration variable accordingly - integrate with respect to y if the boundary is horizontal, and with respect to x if it's vertical. This often simplifies the work.

 

Question 23. Find the area between the curve \( y = \frac{x}{\pi} + 2\sin^2 x \), the axis and the ordinates x = 0 and x = \pi \).
Answer: We need to find the area enclosed by the curve \( y = \frac{x}{\pi} + 2\sin^2 x \), the x-axis, and the vertical lines at \( x = 0 \) and \( x = \pi \). The curve resembles \( y = \sin^2 x \) in shape but with an added linear term. The required area is given by the integral \( \int_0^\pi \left(\frac{x}{\pi} + 2\sin^2 x\right) dx \). We can split this as \( \frac{1}{\pi}\int_0^\pi x \, dx + 2\int_0^\pi \sin^2 x \, dx \). For the first part: \( \frac{1}{\pi} \cdot \frac{x^2}{2}\big|_0^\pi = \frac{\pi}{2} \). For the second part, we use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \), so \( 2\int_0^\pi \sin^2 x \, dx = 2\int_0^\pi \frac{1 - \cos 2x}{2} dx = \int_0^\pi (1 - \cos 2x) \, dx = \left[x - \frac{\sin 2x}{2}\right]_0^\pi = \pi \). Therefore, the total area is \( \frac{\pi}{2} + \pi = \frac{3\pi}{2} \) square units.
In simple words: The curve is made of two parts - a straight rising line and a wavy sine-squared part. To find the total area beneath it, we break the integral into two simpler pieces and solve each separately, then add them together.

Exam Tip: Always split a complex integral into manageable pieces using linearity. Use trigonometric identities (like the double angle formulas) to simplify powers of sine and cosine before integrating.

 

Question 24. Find the area of bounded by the curve y = cos x, the x-axis and the ordinates x = 0 and x = 2\pi \).
Answer: We are given the curve \( y = \cos x \) and need to find the area between it and the x-axis from \( x = 0 \) to \( x = 2\pi \). The cosine function is positive from 0 to \( \frac{\pi}{2} \), negative from \( \frac{\pi}{2} \) to \( \frac{3\pi}{2} \), and positive again from \( \frac{3\pi}{2} \) to \( 2\pi \). Since area is always measured as a positive quantity, we must use absolute values. The total area is the sum of three regions: \( \int_0^{\pi/2} \cos x \, dx + \left|\int_{\pi/2}^{3\pi/2} \cos x \, dx\right| + \int_{3\pi/2}^{2\pi} \cos x \, dx \). The first integral gives \( [\sin x]_0^{\pi/2} = 1 - 0 = 1 \). The second integral gives \( [\sin x]_{\pi/2}^{3\pi/2} = -1 - 1 = -2 \), so its absolute value is 2. The third integral gives \( [\sin x]_{3\pi/2}^{2\pi} = 0 - (-1) = 1 \). Therefore, the total area is \( 1 + 2 + 1 = 4 \) square units.
In simple words: The cosine curve dips below the x-axis in the middle. When finding area, all regions count as positive. We must split the integral at the points where the curve crosses the x-axis and take absolute values to get the true total area.

Exam Tip: Always sketch the curve in the given interval to identify where it crosses the axis. This helps you set up the integral correctly with proper sign handling - use absolute values for regions below the axis.

 

Question 25. Compare the areas under the curves y = cos² x and y = sin² x between x = 0 and x = \pi \).
Answer: We need to find and compare the areas beneath \( y = \cos^2 x \) and \( y = \sin^2 x \) from \( x = 0 \) to \( x = \pi \). For the first curve, the area is \( \int_0^\pi \cos^2 x \, dx \). Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we get \( \int_0^\pi \frac{1 + \cos 2x}{2} dx = \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_0^\pi = \frac{1}{2}\left[\pi + 0 - 0 - 0\right] = \frac{\pi}{2} \). For the second curve, the area is \( \int_0^\pi \sin^2 x \, dx \). Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \), we get \( \int_0^\pi \frac{1 - \cos 2x}{2} dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^\pi = \frac{1}{2}\left[\pi - 0 - 0 + 0\right] = \frac{\pi}{2} \). Both curves enclose the same area, which is \( \frac{\pi}{2} \) square units. This result makes sense because \( \cos^2 x + \sin^2 x = 1 \) for all x, so their graphs are complementary over any interval.
In simple words: Both curves look different - one is highest at the start and decreases, while the other starts low and increases. Yet when we calculate the area under each from 0 to \(\pi\), they turn out to be equal. This happens because they are mirror images of each other in a sense.

Exam Tip: Remember the double angle identities: \( \cos^2 x = \frac{1 + \cos 2x}{2} \) and \( \sin^2 x = \frac{1 - \cos 2x}{2} \). These convert squares of trig functions into forms that are much easier to integrate.

 

Question 26. Using integration, find the area of the triangle, the equations of whose sides are y = 2x + 1, y = 3x + 1 and x = 4.
Answer: We are given three lines that form a triangle: \( y = 2x + 1 \) (side AB), \( y = 3x + 1 \) (side BC), and \( x = 4 \) (side CA). To find the vertices, we solve pairs of equations. For point B, we solve AB and BC: \( 2x + 1 = 3x + 1 \), which gives \( x = 0 \) and \( y = 1 \), so \( B = (0, 1) \). For point C, we solve BC and CA: \( y = 3(4) + 1 = 13 \), so \( C = (4, 13) \). For point A, we solve AB and CA: \( y = 2(4) + 1 = 9 \), so \( A = (4, 9) \). The triangular region can be found by computing the area under each line and using subtraction. The area is \( \int_0^4 (3x + 1) \, dx - \int_0^4 (2x + 1) \, dx = \left[\frac{3x^2}{2} + x\right]_0^4 - \left[\frac{2x^2}{2} + x\right]_0^4 = \left[\frac{3(16)}{2} + 4\right] - \left[\frac{2(16)}{2} + 4\right] = [24 + 4] - [16 + 4] = 28 - 20 = 8 \) square units.
In simple words: Two lines start at the same point on the y-axis but spread apart at different slopes. A vertical line at x = 4 closes off the region to form a triangle. To find the area, we integrate the upper line minus the lower line over the interval from x = 0 to x = 4.

Exam Tip: When working with triangles or polygons defined by line equations, always find the vertices first by solving the equations pairwise. This gives you the bounds and helps verify your setup before integrating.

 

Question 27. Find area of region \(\{(x,y): x^2 \leq y \leq x\}\)
Answer: The region is bounded by the parabola \(y = x^2\) and the line \(y = x\). These two curves intersect where \(x^2 = x\), giving \(x = 0\) and \(x = 1\). At these points, the intersection points are O(0,0) and A(1,1). Since the line \(y = x\) lies above the parabola \(y = x^2\) in the interval [0,1], the required area is calculated by integrating the difference between the line and the parabola.

Area = \(\int_0^1 (x - x^2) \, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\) sq. units

In simple words: The space between a curved parabola and a straight line can be found by subtracting one function from the other and integrating over the interval where they overlap. The result comes out to one-sixth of a square unit.

Exam Tip: Always identify intersection points first, then determine which curve is above the other in the given interval before setting up the integral.

 

Question 28. Find the area of the region bounded by the curve \(y^2=2y-x\) and the y-axis.
Answer: The curve \(y^2 = 2y - x\) can be rewritten by rearranging: \(y^2 - 2y = -x\). Adding 1 to both sides gives \((y-1)^2 = -(x-1)\), which is a parabola with vertex at A(1,1) opening to the left. To find where the parabola meets the y-axis (where \(x = 0\)), substitute: \(y^2 - 2y = 0\), so \(y(y-2) = 0\), giving \(y = 0\) and \(y = 2\). The region is bounded by points O(0,0), B(0,2), and the vertex A(1,1).

From the curve equation, express \(x = 2y - y^2\). The required area is:

Area = \(\int_0^1 x \, dy = \int_0^1 (2y - y^2) \, dy = \left[\frac{2y^2}{2} - \frac{y^3}{3}\right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}\) sq. units

In simple words: A parabola opening sideways meets the vertical axis at two points. The space enclosed between them can be calculated by integrating the x-values with respect to y, giving two-thirds of a square unit.

Exam Tip: When the parabola opens horizontally, integrate with respect to y instead of x to simplify the calculation.

 

Question 29. Draw a rough sketch of the curves \(y=\sin x\) and \(y=\cos x\), as x varies from 0 to \(\frac{\pi}{2}\), and find the area of the region enclosed between them and the x-axis.
Answer: At the given limits, the curves \(y = \cos x\) and \(y = \sin x\) start at different points: at \(x = 0\), \(\cos 0 = 1\) and \(\sin 0 = 0\); at \(x = \frac{\pi}{4}\), both equal \(\frac{1}{\sqrt{2}}\); and at \(x = \frac{\pi}{2}\), \(\cos \frac{\pi}{2} = 0\) and \(\sin \frac{\pi}{2} = 1\). The curves intersect at \(x = \frac{\pi}{4}\). The curve \(y = \cos x\) lies above \(y = \sin x\) from 0 to \(\frac{\pi}{4}\), and \(y = \sin x\) lies above from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\).

The required area is:

\(\text{Area} = \int_0^{\pi/4} \sin x \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx = [-\cos x]_0^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2} = \left(-\frac{1}{\sqrt{2}} + 1\right) + \left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \sqrt{2}\) sq. units

In simple words: Two wavy curves start at different heights and cross in the middle. The area between them is found by splitting the integration at the crossing point and computing each piece separately, which gives 2 minus the square root of 2.

Exam Tip: Always find intersection points of curves and split the integral at those points to ensure correct handling of which curve is on top in each section.

 

Question 30. Find the area of the region bounded by the parabola \(y^2=2x+1\) and the lines \(x-y=1\).
Answer: The parabola \(y^2 = 2x + 1\) can be rewritten as \((y-0)^2 = 2(x + \frac{1}{2})\), showing it has vertex at A(-0.5, 0) and opens rightward. The line \(x - y = 1\) can be written as \(x = y + 1\). To find intersection points, substitute the line equation into the parabola: \(y^2 = 2(y+1) + 1 = 2y + 3\), so \(y^2 - 2y - 3 = 0\). Factoring: \((y-3)(y+1) = 0\), giving \(y = 3\) or \(y = -1\). The corresponding x-values are \(x = 4\) and \(x = 0\), so the curves meet at B(4,3) and C(0,-1). The line meets the x-axis at D(1,0).

Express \(x = \frac{y^2 - 1}{2}\) for the parabola and \(x = y + 1\) for the line. The required area is:

\(\text{Area} = \int_{-1}^{3} (y+1) \, dy - \int_{-1}^{3} \frac{y^2-1}{2} \, dy = \left[\frac{y^2}{2} + y\right]_{-1}^{3} - \frac{1}{2}\left[\frac{y^3}{3} - y\right]_{-1}^{3} = \frac{16}{3}\) sq. units

In simple words: A parabola opening to the right meets a slanted line at two points. The enclosed space between them is found using a vertical slice approach, where you subtract the parabola's contribution from the line's contribution and integrate with respect to y.

Exam Tip: When a region is bounded by curves that are easier to express as functions of y, integrate with respect to y; always express both boundaries as functions of the same variable.

 

Question 31. Find the area of the region bounded by the curve \(y=2x-x^2\) and the straight line \(y=-x\).
Answer: The curve \(y = 2x - x^2\) is a downward-opening parabola. Rewriting it: \(x^2 - 2x = -y\), and adding 1 to both sides: \((x-1)^2 = -(y-1)\), shows the vertex is at B(1,1). The line \(y = -x\) passes through the origin with negative slope. To find intersection points, set \(2x - x^2 = -x\), which gives \(x^2 - 3x = 0\), so \(x = 0\) or \(x = 3\). At these x-values, \(y = 0\) and \(y = -3\) respectively, so the curves meet at O(0,0) and A(3,-3). A perpendicular from the parabola's highest point to the x-axis meets at C(2,0).

The required area is:

\(\text{Area} = \int_0^3 (2x - x^2) \, dx - \int_0^3 (-x) \, dx = \int_0^3 (3x - x^2) \, dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}\) sq. units

In simple words: A curved parabola and a straight line intersect at two points. The area enclosed between them is calculated by integrating the difference of their expressions over the interval from one intersection point to the other.

Exam Tip: To find enclosed area between a parabola and a line, determine intersection points first, then integrate the difference (upper function minus lower function) over the appropriate interval.

 

Question 32. Find the area of the region bounded by the curve \((y-1)^2=4(x+1)\) and the line \(y=x-1\).
Answer: The equation \((y-1)^2 = 4(x+1)\) represents a parabola opening rightward with vertex at B(-1,1). Rewriting, \(x = \frac{(y-1)^2}{4} - 1\). The line is \(y = x - 1\), which can be expressed as \(x = y + 1\). To find intersections, substitute: \((y-1)^2 = 4(y+1+1) = 4(y+2)\), so \((y-1)^2 = 4y + 8\), giving \(y^2 - 2y + 1 = 4y + 8\), thus \(y^2 - 6y - 7 = 0\). Solving: \((y-7)(y+1) = 0\), so \(y = 7\) or \(y = -1\). The corresponding x-values are \(x = 8\) and \(x = 0\), giving intersection points D(8,7) and E(0,-1).

Express \(x = \frac{(y-1)^2}{4} - 1\) for the parabola. The required area is:

\(\text{Area} = \int_{-1}^{7} (y+1) \, dy - \int_{-1}^{7} \left[\frac{(y-1)^2}{4} - 1\right] dy = \frac{64}{3}\) sq. units

In simple words: A sideways-opening parabola and a diagonal line bound a region between two intersection points. Integrating the line's x-value minus the parabola's x-value with respect to y gives the total enclosed area.

Exam Tip: For parabolas opening left or right, always express x as a function of y and integrate with respect to y for cleaner calculations.

 

Question 33. Find the area of the region bounded by the curve \(y=\sqrt{x}\) and the line \(y=x\).
Answer: The curve \(y = \sqrt{x}\) or \(y^2 = x\) is a rightward-opening parabola with vertex at the origin O(0,0). The line \(y = x\) passes through the origin at 45 degrees. Setting them equal: \(\sqrt{x} = x\), so \(x = x^2\), giving \(x^2 - x = 0\), thus \(x = 0\) or \(x = 1\). The curves intersect at O(0,0) and A(1,1). Since \(y = \sqrt{x}\) is above \(y = x\) in the interval [0,1], the enclosed area is:

\(\text{Area} = \int_0^1 (\sqrt{x} - x) \, dx = \left[\frac{2x^{3/2}}{3} - \frac{x^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\) sq. units

In simple words: A square root curve starts above a diagonal line and they meet at the corner point (1,1). The space between them is found by integrating the difference of the two functions from 0 to 1.

Exam Tip: Always check which curve is above the other by testing a point in the interval; this ensures the integrand is positive and the final answer is correct.

 

Question 34. Find the area of the region included between the parabola \(y^2=3x\) and the circle \(x^2+y^2-6x=0\), lying in the first quadrant.
Answer: Rewrite the circle equation as \((x-3)^2 + y^2 = 9\), which has center at (3,0) and radius 3. The parabola \(y^2 = 3x\) has vertex at the origin and opens rightward. To find intersection points, substitute \(y^2 = 3x\) into the circle equation: \(x^2 + 3x - 6x = 0\), so \(x^2 - 3x = 0\), giving \(x = 0\) or \(x = 3\). When \(x = 3\), \(y = \pm 3\); when \(x = 0\), \(y = 0\). The relevant first-quadrant points are O(0,0) and A(3,3), with C(3,0) on the x-axis.

For the circle: \(y = \sqrt{9-(x-3)^2}\), and for the parabola: \(y = \sqrt{3x}\). The required area is:

\(\text{Area} = \int_0^3 \sqrt{9-(x-3)^2} \, dx - \int_0^3 \sqrt{3x} \, dx = \frac{9\pi}{4} - 2\sqrt{3}\) sq. units

In simple words: A circle and a parabola overlap in the first quadrant, meeting at two points. The region between them (circle on top, parabola below) is found by integrating the circle's height minus the parabola's height.

Exam Tip: When dealing with circular boundaries, use substitution \(x = a + r\sin\theta\) to simplify the square root integral, or apply the standard formula for circular segments.

 

Question 35. Find the area bounded by the curve \(y=\cos x\) between \(x=0\) to \(x=2\pi\).
Answer: The curve \(y = \cos x\) is a periodic function that oscillates between -1 and 1. Over the interval [0, 2π], it follows a specific pattern: from 0 to π/2, it decreases from 1 to 0; from π/2 to π, it decreases from 0 to -1; from π to 3π/2, it increases from -1 to 0; and from 3π/2 to 2π, it increases from 0 to 1. Since the curve dips below the x-axis from π/2 to 3π/2, that portion contributes a negative value to a direct integral. To find the total enclosed area (counting all regions as positive), split the integral:

\(\text{Area} = \left|\int_0^{\pi/2} \cos x \, dx\right| + \left|\int_{\pi/2}^{3\pi/2} \cos x \, dx\right| + \left|\int_{3\pi/2}^{2\pi} \cos x \, dx\right| = 1 + 2 + 1 = 4\) sq. units

In simple words: A cosine wave completes one full cycle from 0 to 2π, going above and below the x-axis. To find the total area enclosed between the curve and the axis, take the absolute value of each section and add them together.

Exam Tip: When a curve crosses the x-axis within the integration limits, always split the integral at those crossing points and use absolute values to ensure all enclosed areas are counted as positive quantities.

 

Question 36. Using integration, find the area of the region in the first quadrant, enclosed by the x-axis, the line y=x and the circle x²+y²=32
Answer: The boundaries of the area to find include the circle x² + y² = 32, the line y = x, and the region must lie in the first quadrant.

From the circle equation x² + y² = 32, the center sits at the origin and the radius measures 4√2 units.

To locate where the circle and line meet, substitute y = x into the circle equation:
x² + x² = 32
2x² = 32
x² = 16
x = ± 4

Substituting back into the line equation: y = ± 4. In the first quadrant, the intersection point is A = (4,4).

Since both variables have even powers in the circle equation, the curve is symmetric about both axes. Rewrite the circle as y = √(32 - x²).

Drop a perpendicular from point A to the x-axis, meeting it at C = (4, 0). The required area equals the region OADO.

Split this into two parts:

Area of OADO = Area of OAC + Area of CADC

\[ \text{Area of OADO} = \int_{4\sqrt{2}}^{4} \sqrt{32 - x^2} \, dx + \int_{0}^{4} x \, dx \]

For the first integral, use the formula \( \int \sqrt{a^2 - x^2} \, dx = \frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) \) and \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \):

\[ = \left[\frac{x\sqrt{32 - x^2}}{2} + \frac{32}{2} \sin^{-1}\left(\frac{x}{4\sqrt{2}}\right)\right]_{4\sqrt{2}}^{4} + \left[\frac{x^2}{2}\right]_{0}^{4} \]

\[ = \left\{\frac{(4)\sqrt{32 - 16}}{2} + \frac{32}{2} \sin^{-1}\left(\frac{4}{4\sqrt{2}}\right)\right\} - \left\{\frac{(4\sqrt{2})\sqrt{32 - 32}}{2} + \frac{32}{2} \sin^{-1}(1)\right\} + \left[\frac{16}{2}\right] \]

\[ = \left\{\frac{(4)(4)}{2} + \frac{32}{2} \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\right\} - \{0 + 16 \cdot 90°\} + 8 \]

Since \( \sin^{-1}(1) = 90° \) and \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45° \):

\[ = \{8 + 16(45°)\} - \{16(0)\} + 8 = 8 + 4\pi - 8 = 4\pi \]

The area of the required region is 4π sq. units.
In simple words: Find where the line and circle intersect in the first quadrant. Split the curved region into a piece under the circle arc and a triangular piece. Use integration formulas for both parts and add them together.

Exam Tip: Always identify intersection points carefully and use symmetry properties to simplify. Recognize inverse sine values at standard angles (sin⁻¹(1) = 90°, sin⁻¹(1/√2) = 45°) to avoid arithmetic errors.

 

Question 37. Using integration, find the area of the triangle whose vertices are A(2,3), B(4,7) and C(6,2).
Answer: Given the three vertices A(2,3), B(4,7), and C(6,2), the triangle area may be found by calculating the area under each side and combining them appropriately.

For side AB, use the two-point line formula to find the equation:
\[ \frac{y - 3}{7 - 3} = \frac{x - 2}{4 - 2} \]
\[ \frac{y - 3}{4} = \frac{x - 2}{2} \]
\[ y - 3 = 2(x - 2) \]
\[ y = 2x - 1 \]

Area under line AB (from x = 2 to x = 4):
\[ \text{Area of ABDE} = \int_{2}^{4} (2x - 1) \, dx = \left[x^2 - x\right]_{2}^{4} \]
\[ = [16 - 4] - [4 - 2] = 12 - 2 = 10 \text{ sq. units} \quad \text{...(1)} \]

For side BC, using the two-point formula with B(4,7) and C(6,2):
\[ \frac{y - 7}{2 - 7} = \frac{x - 4}{6 - 4} \]
\[ \frac{y - 7}{-5} = \frac{x - 4}{2} \]
\[ y = \frac{34 - 5x}{2} \]

Area under line BC (from x = 4 to x = 6):
\[ \text{Area of BCFE} = \int_{4}^{6} \frac{34 - 5x}{2} \, dx = \frac{1}{2}\left[34x - \frac{5x^2}{2}\right]_{4}^{6} \]
\[ = \frac{1}{2}\left[\left(204 - 90\right) - \left(136 - 40\right)\right] = \frac{1}{2}[114 - 96] = 9 \text{ sq. units} \quad \text{...(2)} \]

For side CA, using the two-point formula with C(6,2) and A(2,3):
\[ \frac{y - 2}{3 - 2} = \frac{x - 6}{2 - 6} \]
\[ \frac{y - 2}{1} = \frac{x - 6}{-4} \]
\[ y = \frac{14 - x}{4} \]

Area under line CA (from x = 2 to x = 6):
\[ \text{Area of ACFD} = \int_{2}^{6} \frac{14 - x}{4} \, dx = \frac{1}{4}\left[14x - \frac{x^2}{2}\right]_{2}^{6} \]
\[ = \frac{1}{4}\left[\left(84 - 18\right) - \left(28 - 2\right)\right] = \frac{1}{4}[66 - 26] = 10 \text{ sq. units} \quad \text{...(3)} \]

When areas under all three sides are combined on a graph, the area under side AC overlaps with the sum of areas under AB and BC. Therefore, the triangle area is:

\[ \text{Area of triangle ABC} = \text{Area under AB} + \text{Area under BC} - \text{Area under AC} \]
\[ = 10 + 9 - 10 = 9 \text{ sq. units} \]
In simple words: Find the straight-line equation for each side of the triangle. Integrate each equation between its x-coordinates. Combine the three integrals by adding two outer sides and subtracting the inner side to get the triangle's enclosed area.

Exam Tip: Always verify that the line equations are correct by checking that both vertices satisfy the equation. Be careful with the overlapping region - subtract the area under the side connecting the first and last vertices.

 

Question 38. Using integration, find the area of the triangle whose vertices are A(1,3), B(2,5) and C(3,4).
Answer: Given the three vertices A(1,3), B(2,5), and C(3,4), determine the triangle area by finding the equation of each side and integrating.

For side AB, apply the two-point line formula with A(1,3) and B(2,5):
\[ \frac{y - 3}{5 - 3} = \frac{x - 1}{2 - 1} \]
\[ \frac{y - 3}{2} = \frac{x - 1}{1} \]
\[ y = 2x + 1 \]

Area under line AB (from x = 1 to x = 2):
\[ \text{Area of ABDE} = \int_{1}^{2} (2x + 1) \, dx = \left[x^2 + x\right]_{1}^{2} \]
\[ = [4 + 2] - [1 + 1] = 6 - 2 = 4 \text{ sq. units} \quad \text{...(1)} \]

For side BC, using the two-point formula with B(2,5) and C(3,4):
\[ \frac{y - 5}{4 - 5} = \frac{x - 2}{3 - 2} \]
\[ \frac{y - 5}{-1} = \frac{x - 2}{1} \]
\[ y = 7 - x \]

Area under line BC (from x = 2 to x = 3):
\[ \text{Area of BCFE} = \int_{2}^{3} (7 - x) \, dx = \left[7x - \frac{x^2}{2}\right]_{2}^{3} \]
\[ = \left[21 - \frac{9}{2}\right] - \left[14 - 2\right] = \frac{33}{2} - 12 = \frac{9}{2} \text{ sq. units} \quad \text{...(2)} \]

For side CA, using the two-point formula with C(3,4) and A(1,3):
\[ \frac{y - 4}{3 - 4} = \frac{x - 3}{1 - 3} \]
\[ \frac{y - 4}{-1} = \frac{x - 3}{-2} \]
\[ y = \frac{x + 5}{2} \]

Area under line CA (from x = 1 to x = 3):
\[ \text{Area of ACFD} = \int_{1}^{3} \frac{x + 5}{2} \, dx = \frac{1}{2}\left[\frac{x^2}{2} + 5x\right]_{1}^{3} \]
\[ = \frac{1}{2}\left[\left(\frac{9}{2} + 15\right) - \left(\frac{1}{2} + 5\right)\right] = \frac{1}{2}\left[\frac{39}{2} - \frac{11}{2}\right] = 7 \text{ sq. units} \quad \text{...(3)} \]

The area under side CA overlaps with the combined areas under sides AB and BC. Thus, the triangle's area becomes:

\[ \text{Area of triangle ABC} = \text{Area under AB} + \text{Area under BC} - \text{Area under CA} \]
\[ = 4 + \frac{9}{2} - 7 = \frac{9}{2} - 3 = \frac{3}{2} \text{ sq. units} \]
In simple words: Write the equation for each of the three sides. Integrate each equation over its corresponding x-range. The triangle area is found by adding the two sides that form the base and subtracting the slanted side that crosses over them.

Exam Tip: Double-check the line equations by substituting both vertex coordinates. Watch for fractional results - verify arithmetic carefully, especially when combining fractions.

 

Question 39. Using integration, find the area of the triangular region bounded by the lines y=2x+1, y=3x+1 and x=4.
Answer: The triangular region is bounded by three lines: y = 2x+1, y = 3x+1, and x = 4. Find the vertices first by solving pairs of these equations.

Intersection of lines y = 2x+1 and y = 3x+1:
2x + 1 = 3x + 1
x = 0
Substituting into y = 2x+1: y = 1
Vertex B = (0,1)

Intersection of lines y = 3x+1 and x = 4:
y = 3(4) + 1 = 13
Vertex C = (4,13)

Intersection of lines y = 2x+1 and x = 4:
y = 2(4) + 1 = 9
Vertex A = (4,9)

These points set the upper and lower bounds for integration. The triangular region lies between the two lines from x = 0 to x = 4.

For line AB with equation y = 2x+1, the area under it is:
\[ \text{Area under AB} = \int_{0}^{4} (2x + 1) \, dx = \left[x^2 + x\right]_{0}^{4} \]
\[ = [16 + 4] - [0 + 0] = 20 \text{ sq. units} \quad \text{...(1)} \]

For line BC with equation y = 3x+1, the area under it is:
\[ \text{Area under BC} = \int_{0}^{4} (3x + 1) \, dx = \left[\frac{3x^2}{2} + x\right]_{0}^{4} \]
\[ = \left[24 + 4\right] - [0 + 0] = 28 \text{ sq. units} \quad \text{...(2)} \]

The triangle's area is the difference between these two regions because the steeper line (BC) encloses more area beneath it. Removing the area under the gentler line (AB) leaves the triangular region:

\[ \text{Area of triangle ABC} = \text{Area under BC} - \text{Area under AB} \]
\[ = 28 - 20 = 8 \text{ sq. units} \]
In simple words: Locate the three vertices by solving pairs of line equations. Integrate both non-vertical lines between x = 0 and x = 4. The triangle sits between these two curves, so subtract the smaller area from the larger area.

Exam Tip: When a region is bounded by two slanted lines and a vertical line, integrate both slanted lines over the same x-interval and subtract. This approach naturally gives the area trapped between them.