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CIE A-Level Maths Study Notes

1.8.3 Evaluating Definite Integrals

Evaluating definite integrals is a fundamental concept under the topic of Integration. This section focuses on methods to evaluate definite integrals, with an emphasis on dealing with improper integrals. Improper integrals occur when the limits of integration are infinite or when the integrand (the function to be integrated) is discontinuous at certain points.

Understanding Definite Integrals

A definite integral, denoted as abf(x)dx\int_{a}^{b} f(x) \, dx, where f(x)f(x) is the integrand, and aa and bb are the limits of integration, calculates the net area under the curve of f(x)f(x) from x=ax = a to x=bx = b.

  • Limits of Integration: These values, aa and bb, define the interval over which the function is integrated.
  • Net Area: The integral calculates the 'net' area, considering areas above the x-axis as positive and below as negative.
definite integral

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Techniques for Evaluating Definite Integrals

Evaluating a definite integral involves finding the antiderivative of the function and then applying the limits of integration.

1. Find the Antiderivative

Determine the indefinite integral or antiderivative of f(x)f(x).

2. Apply the Fundamental Theorem of Calculus

Use the formula abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) \, dx = F(b) - F(a), where FF is the antiderivative of f(x)f(x).

Example: Evaluating a Basic Definite Integral

Consider the integral 01x2dx\int_{0}^{1} x^2 \, dx:

1. Find the antiderivative of x2x^2, which is 13x3\frac{1}{3}x^3, using the power rule.

2. Apply the Fundamental Theorem of Calculus:

Evaluate 13x3\frac{1}{3}x^3 from 0 to 1, resulting in 13(1)313(0)3\frac{1}{3}(1)^3 - \frac{1}{3}(0)^3.

3. Simplify to get 13\frac{1}{3}.

This represents the area under the curve x2x^2 from x=0x = 0 to x=1x = 1.

Improper Integrals

improper integral

Image courtesy of wikipedia

Improper integrals arise in two primary scenarios:

1. Infinite Limits of Integration:

Either or both limits of a definite integral are infinite (positive or negative).

infinite limits

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2. Undefined Integrand at a Limit:

The function to be integrated is undefined at one or both limits of integration.

Handling Improper Integrals

To evaluate these integrals:

  • Replace the problematic limit with a variable that approaches the problematic value (such as infinity or zero).
  • Evaluate the integral with this new limit.
  • Finally, consider the behaviour of the integral as the variable approaches its limit.

Examples

Example 1: Undefined Integrand at a Limit

Evaluate: 021xdx\int_0^2 \frac{1}{\sqrt{x}} \, dx

Solution:

Here, the integrand is undefined at x=0x = 0. Substitute 00 with a variable aa, approaching zero:

a2x12dx=[x12+112+1]a2\int_a^2 x^{-\frac{1}{2}} \, dx = \left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_a^2=[2x12]a2= \left[2x^{\frac{1}{2}}\right]_a^2

At x=2x = 2, the value is 222 \sqrt{2}. As aa approaches zero, the value approaches 00.

Thus, the integral is 220=222\sqrt{2} - 0 = 2\sqrt{2}.

021xdx=22\therefore \int_0^2 \frac{1}{\sqrt{x}} \, dx = 2\sqrt{2}

Example 2: Infinite Limit of Integration

Evaluate: 214xdx \int_2^{\infty} \frac{1}{4\sqrt{x}} \, dx

Solution:

Here, the upper limit is infinite. Consider an upper limit bb, approaching infinity:

2bx124dx=[x12+14(12+1)]2b\int_2^b \frac{x^{-\frac{1}{2}}}{4} \, dx = \left[\frac{x^{-\frac{1}{2}+1}}{4\left(-\frac{1}{2}+1\right)}\right]_2^b =[x2]2b= \left[\frac{\sqrt{x}}{2}\right]_2^b

As bb approaches infinity, the value of x2\frac{\sqrt{x}}{2} also approaches infinity.

Thus, the integral is 22=\infty - \frac{\sqrt{2}}{2} = \infty.

214xdx=\therefore \int_2^{\infty} \frac{1}{4\sqrt{x}} \, dx = \infty

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