Evaluating definite integrals

2 Evaluating definite integrals

When you evaluate a definite integral the result will usually be a number. To see how to evaluate a definite integral consider the following Example.

Example 9

Find the definite integral of $x^{2}$ from $1$ to $4$ ; that is, find $\int_{1}^{4} x^{2} d x$

Solution

$\int x^{2} d x = \frac{1}{3} x^{3} + c$

Here $f (x) = x^{2}$ and $F (x) = \frac{x^{3}}{3}$ . Thus, according to our definition

$\int_{1}^{4} x^{2} d x = F (4) - F (1) = \frac{4^{3}}{3} - \frac{1^{3}}{3} = 21$

Writing $F (b) - F (a)$ each time we calculate a definite integral becomes laborious so we replace this difference by the shorthand notation ${[F (x)]}_{a}^{b}$ . Thus

${[F (x)]}_{a}^{b} \equiv F (b) - F (a)$

Thus, from now on, we shall write

$\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b}$

so that, for example

$\int_{1}^{4} x^{2} d x = {[\frac{x^{3}}{3}]}_{1}^{4} = \frac{4^{3}}{3} - \frac{1^{3}}{3} = 21$

Example 10

Find the definite integral of $cos x$ from $0$ to $\frac{π}{2}$ ; that is, find $\int_{0}^{π ∕ 2} cos x d x$ .

Solution

Since $\int cos x d x = sin x + c$ then

\begin{array}{rcl} \int_{0}^{π ∕ 2} cos x d x & = & {[sin x]}_{0}^{π ∕ 2} \\ = & sin (\frac{π}{2}) - sin 0 = 1 - 0 = 1 \end{array}

Always remember, that if you use a calculator to evaluate any trigonometric functions, you must work in radian mode .

Task!

Find the definite integral of $x^{2} + 1$ from $1$ to $2$ ; that is; find $\int_{1}^{2} (x^{2} + 1) d x$

First perform the integration:

${[\frac{1}{3} x^{3} + x]}_{1}^{2}$ . Now insert the limits of integration, the upper limit first, and hence evaluat the integral:

$(\frac{8}{3} + 2) - (\frac{1}{3} + 1) = \frac{10}{3}$ or 3.333 (3 d.p.).

Task!

Find $\int_{2}^{1} (x^{2} + 1) d x$ .

This Task is very similar to the previous Task. Note the limits have been interchanged:

${[\frac{1}{3} x^{3} + x]}_{2}^{1} = [\frac{1}{3} + 1] - [\frac{8}{3} + 2] = - \frac{10}{3}$ .

Note from these two Tasks that interchanging the limits of integration, changes the sign of the answer.

Key Point 3

If you interchange the limits, you must change the sign:

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x

When a spring is fixed at one end and stretched at the free end it exerts a restoring force that is proportional to the displacement of the free end. The constant of proportionality $k N m^{- 1}$ is known as the stiffness of the spring. Calculate the work done in stretching a spring with stiffness $k$ from displacement $x_{1}$ m to displacement $x_{2}$ m ( $x_{2} > x_{1}$ ) given that the work done $(W)$ is the product of force and displacement.

The restoring force varies during the displacement. So the work done during the extension cannot be determined from a single simple product.

Consider a small element $Δ x$ of the extension beyond an arbitrary displacement $x$ . The element is sufficiently small that the force during the displacement can be regarded as constant and equal to the force at displacement $x$ is $k x$ . So the work done $Δ W$ in extending the spring from displacement $x$ to displacement $x + Δ x$ is approximately $k x Δ x$ .

Using the idea of integration as a limit of a sum, in this case as $Δ x$ tends to zero,

$W = \int_{x_{1}}^{x_{2}} k x d x = {[\frac{1}{2} k x^{2}]}_{x_{1}}^{x_{2}} = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$

Exercises

Evaluate
1. $\int_{0}^{1} x^{2} d x$ ,
2. $\int_{2}^{3} \frac{1}{x^{2}} d x$
3. $\int_{1}^{2} e^{x} d x$
4. $\int_{- 1}^{1} (1 + t^{2}) d t$
Find
1. $\int_{0}^{π ∕ 3} cos 2 x d x$
2. $\int_{0}^{π} sin x d x$
3. $\int_{1}^{3} e^{2 t} d t$

1. $\frac{1}{3}$
2. $\frac{1}{6}$
3. $e^{2} - e^{1} = 4.671$
4. 2.667
1. $\sqrt{3} ∕ 4 = 0.4330$
2. 2
3. 198.019

2 Evaluating definite integrals

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