Maharashtra Board Class 12 Maths Commerce Part I Chapter 6 Definite Integration PDF Download

Definite Integration

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Definite Integral

Properties of Definite Integral

Introduction

We know that if f(x) is a continuous function of x, then there exists a function I(x) such that I'(x) = f(x). In this case, I(x) is an integral of f(x) with respect to x and we denote it by \(\int f(x) \, dx = I(x) + c\).

Now if we restrict the domain of f(x) to (a, b), then the difference I(b) - I(a) is called definite integral of f(x) with respect to x on the interval [a, b] and is denoted by \(\int_a^b f(x) \, dx\).

Thus \(\int_a^b f(x) \, dx = I(b) - I(a)\)

The numbers a and b are called limits of integration. a is referred to as the lower limit of integral and b is the upper limit of integral.

Note that the domain of the variable x is restricted to the interval (a, b) and a, b are finite numbers.

Teacher's Note

Think of definite integral like measuring the area under a curve. If you want to find how much water a curved container holds between two points, that is a definite integral.

Exam Trick

Remember: The constant c always disappears in definite integrals! When you put the limits, the c cancels out, so you do not need to write it.

Points to Remember

Definite integral has two limits: lower and upper.

The answer is always a number, not a function.

The constant of integration c is not needed because it cancels out.

Let's Learn

Fundamental Theorem of Integral Calculus

Let f be a continuous function defined on (a, b).

\(\int f(x) \, dx = \phi(x) + c\)

Then \(\int_a^b f(x) \, dx = [\phi(x) + c]_a^b = [\phi(b) + c] - [\phi(a) + c] = \phi(b) - \phi(a)\)

There is no need of taking the constant of integration c because it gets eliminated.

Teacher's Note

This theorem is very important. It tells us how to calculate definite integrals using antiderivatives. It is like finding the total distance by looking at the starting and ending positions.

Exam Trick

Always find the antiderivative first, then put the upper limit, then put the lower limit, and subtract. Follow this order carefully.

Points to Remember

Find the antiderivative I(x) of the function.

Calculate I(b) by putting x = b.

Calculate I(a) by putting x = a.

Subtract I(a) from I(b).

Solved Examples

Ex 1: Evaluate

i) \(\int_2^3 x^4 \, dx\)

ii) \(\int_0^1 \frac{1}{2x+5} \, dx\)

iii) \(\int_0^1 \frac{1}{\sqrt{1+x} + \sqrt{x}} \, dx\)

Solution:

i) Here f(x) = x^4, \(\phi(x) = \frac{x^5}{5} + c\)

\(\int_2^3 f(x) \, dx = [\phi(x)]_2^3\)

\(\int_2^3 x^4 \, dx = \left[\frac{x^5}{5}\right]_2^3 = \frac{3^5}{5} - \frac{2^5}{5} = \frac{243}{5} - \frac{32}{5} = \frac{211}{5}\)

ii) \(\int_0^1 \frac{1}{2x+5} \, dx = \frac{1}{2} [\log|2x+5|]_0^1 = \frac{1}{2} [\log 7 - \log 5] = \frac{1}{2} \log \frac{7}{5}\)

iii) \(\int_0^1 \frac{1}{\sqrt{1+x} + \sqrt{x}} \, dx = \int_0^1 \frac{\sqrt{1+x} - \sqrt{x}}{(\sqrt{1+x} + \sqrt{x})(\sqrt{1+x} - \sqrt{x})} \, dx = \int_0^1 \frac{\sqrt{1+x} - \sqrt{x}}{1+x-x} \, dx = \int_0^1 (\sqrt{1+x} - \sqrt{x}) \, dx\)

Teacher's Note

When you see complicated expressions in definite integrals, try rationalizing the denominator. This makes the problem much easier, just like simplifying a fraction.

Exam Trick

Remember: After you find the antiderivative, always put the upper limit first, then subtract the lower limit. Do not reverse this order.

Points to Remember

The definite integral is always a fixed number.

The constant c disappears because I(b) + c and I(a) + c cancel.

Always simplify before integrating when possible.

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