Worked examples · Sheet 3

Definite integrals and the fundamental theorem of calculus

Theory used: Fundamental formula of integral calculus Integral and area Properties of the definite integral Integral mean value theorem Integral function Fundamental theorem of integral calculus

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute

\int_1^4 \left( 3\sqrt{x} - \frac{1}{x^2} \right) dx .

\int \left( 3\,x^{\frac{1}{2}} - x^{-2} \right) dx = 3 \cdot \frac{2}{3}\,x^{\frac{3}{2}} + x^{-1} + c = 2\,x^{\frac{3}{2}} + \frac{1}{x} + c

\begin{aligned} \int_1^4 \left( 3\sqrt{x} - \frac{1}{x^2} \right) dx &= \left[ 2\,x^{\frac{3}{2}} + \frac{1}{x} \right]_1^4 = \left( 2 \cdot 8 + \frac{1}{4} \right) - (2 + 1) \\[6pt] &= \frac{65}{4} - 3 = \frac{53}{4} \end{aligned}

By the fundamental formula of calculus, an antiderivative of the integrand, which is continuous on $[1, 4]$ , is enough. Writing the root and the reciprocal as powers, the antiderivative is found with the power formula; the constant $c$ can be omitted, because it cancels in the difference. In the computation $4^{\frac{3}{2}} = \left( \sqrt{4} \right)^3 = 8$ . We subtract the value at $1$ from the value at $4$ , and obtain $\dfrac{53}{4}$ .

Exercise 2

Compute $\displaystyle\int_0^3 \left( x^2 - 2x \right) dx$ and the area of the region between the graph of $f(x) = x^2 - 2x$ and the $x$ -axis for $0 \leq x \leq 3$ .

\int_0^3 \left( x^2 - 2x \right) dx = \left[ \frac{x^3}{3} - x^2 \right]_0^3 = 9 - 9 = 0

x^2 - 2x = x\,(x - 2) < 0 \ \text{ for } 0 < x < 2, \qquad x^2 - 2x > 0 \ \text{ for } 2 < x \leq 3

\int_0^2 \left( 2x - x^2 \right) dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}

\int_2^3 \left( x^2 - 2x \right) dx = \left[ \frac{x^3}{3} - x^2 \right]_2^3 = 0 - \left( \frac{8}{3} - 4 \right) = \frac{4}{3}

A = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}

y = x^2 - 2x

The two regions between

y = x^{2} - 2 x

and the

x

-axis both have area

\frac{4}{3}

, so the integral on

[0, 3]

equals

0

and the area equals

\frac{8}{3}

The integral is computed with the fundamental formula and equals $0$ . This does not mean that the area is zero, because the integral counts with a minus sign the parts of the region that lie below the axis. The function $x\,(x - 2)$ is negative between the two zeros $0$ and $2$ and positive for $x > 2$ . We therefore split the interval at $x = 2$ and, by additivity, compute the area of each part with the integral of the positive function: on the first piece $2x - x^2$ , on the second $x^2 - 2x$ . The two areas both equal $\dfrac{4}{3}$ , so the integral, which is their difference, is $0$ , and the area, which is their sum, is $\dfrac{8}{3}$ .

Exercise 3

Compute the integral mean of $f(x) = 6x - 3x^2$ on the interval $[0, 2]$ and find the points $c \in [0, 2]$ at which $f(c)$ is equal to the mean.

\frac{1}{2 - 0} \int_0^2 \left( 6x - 3x^2 \right) dx = \frac{1}{2} \Big[ 3x^2 - x^3 \Big]_0^2 = \frac{1}{2}\,(12 - 8) = 2

6c - 3c^2 = 2 \iff 3c^2 - 6c + 2 = 0 \iff c = \frac{6 \pm \sqrt{36 - 24}}{6} = 1 \pm \frac{\sqrt{3}}{3}

c_1 = 1 - \frac{\sqrt{3}}{3} \approx 0.42, \qquad c_2 = 1 + \frac{\sqrt{3}}{3} \approx 1.58

The integral mean is the integral divided by the width of the interval, here $2$ , and it equals $2$ . The function is continuous on $[0, 2]$ , so the integral mean value theorem guarantees that there exists at least one point at which $f$ takes the value $2$ . To find it we solve the second-degree equation $f(c) = 2$ . The discriminant is $36 - 24 = 12$ and $\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt{3}}{6} = \dfrac{\sqrt{3}}{3}$ . Both solutions, about $0.42$ and $1.58$ , lie in $[0, 2]$ , therefore the required points are two; the theorem ensures the existence of at least one point, and not its uniqueness.

Exercise 4

Let

F(x) = \int_0^x \left( t^2 - 4t + 3 \right) e^{-t}\,dt, \qquad x \in \mathbb{R} .

Compute $F(0)$ and $F'(x)$ , find the intervals on which $F$ increases and decreases, and its relative maximum and minimum points.

F(0) = \int_0^0 \left( t^2 - 4t + 3 \right) e^{-t}\,dt = 0

F'(x) = \left( x^2 - 4x + 3 \right) e^{-x} = (x - 1)(x - 3)\,e^{-x}

	$x < 1$	$1 < x < 3$	$x > 3$
$x - 1$	$-$	$+$	$+$
$x - 3$	$-$	$-$	$+$
$F'(x)$	$+$	$-$	$+$
$F$	increasing	decreasing	increasing

x = 1 \ \text{ relative maximum point}, \qquad x = 3 \ \text{ relative minimum point}

An integral with equal bounds equals $0$ , so $F(0) = 0$ . The integrand is continuous on $\mathbb{R}$ , therefore by the fundamental theorem of integral calculus the integral function is differentiable and its derivative is the integrand evaluated at $x$ ; there is no need to compute the integral. The factor $e^{-x}$ is always positive, so the sign of $F'$ is that of the polynomial $(x - 1)(x - 3)$ , positive outside $[1, 3]$ and negative inside. By the criterion of the sign of the derivative and monotonicity, $F$ is increasing on $(-\infty, 1]$ and on $[3, +\infty)$ and decreasing on $[1, 3]$ . Consequently $x = 1$ , where $F$ goes from increasing to decreasing, is a relative maximum point, and $x = 3$ is a relative minimum point.

As a check, an antiderivative of the integrand is $-(t - 1)^2\,e^{-t}$ , as one verifies by differentiating; therefore $F(x) = 1 - (x - 1)^2\,e^{-x}$ , with $F(1) = 1$ and $F(3) = 1 - 4e^{-3} \approx 0.80$ .