## Section3.1Average Value and Area Revisited

In this section we apply the tool of integration to two applications, namely that of finding the average value of a function and that of determining the area of a region bounded by functions.

### Subsection3.1.1Average Value of a Function

The average of some finite set of values is a familiar concept. If, for example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6, 3, 2, 7, 8, then the average score is the sum of these numbers divided by the size of the class:

\begin{equation*} \hbox{average score} = {10+ 9+ 10+ 8+ 7+ 5+ 7+ 6+ 3+ 2+ 7+ 8\over 12}={82\over 12}\approx 6.83\text{.} \end{equation*}

Although the above was a discrete example, we can extend the idea to that of continuous functions. If we are given an integrable function $f$ on a closed interval $[a,b]\text{,}$ then the average value of $f$ on $[a,b]$ can be defined in a similar fashion. We begin by dividing the interval $[a,b]$ into $n$ subintervals with equal width

\begin{equation*} \Delta x = \frac{b-a}{n} \end{equation*}

and partition

\begin{equation*} P=\{x_1,x_2,\dots,x_n\} \end{equation*}

such that $x_i$ is in the $i$-th subinterval. Then the average value $f_{avg}$ over the $n$ subintervals and choices $x_i$ is given by

\begin{equation*} f_{avg} = \frac{f(x_1)+f(x_2)+\dots+f(x_n)}{n} = \frac{\ds\sum_{i=1}^n f(x_i)}{n}\text{.} \end{equation*}

We now use a typical trick in mathematics, namely we multiply both sides by 1 in the form of $\frac{b-a}{b-a}$ and rearrange terms:

\begin{equation*} f_{avg} = \frac{b-a}{b-a} \cdot \frac{\ds\sum_{x=1}^n f(x_i)}{n} = \frac{b-a}{n}\cdot \frac{\ds\sum_{i=1}^n f(x_i)}{b-a} \end{equation*}

Next, we replace $\frac{b-a}{n}$ by $\Delta x$ to obtain

\begin{equation*} f_{avg} = \Delta x \cdot\frac{\sum_{i=1}^n f(x_i)}{b-a} = \frac{1}{b-a} \cdot \sum_{i=1}^n f(x_i)\Delta x\text{.} \end{equation*}

Taking the limit as $n\to\infty\text{,}$ we finally have

\begin{equation*} \begin{split} f_{avg} \amp = \lim_{n\to\infty} \,\frac{1}{b-a} \cdot \sum_{i=1}^n f(x_i)\Delta x \\ \amp = \frac{1}{b-a}\, \lim_{n\to\infty}\,\sum_{i=1}^n f(x_i)\Delta x\\ \amp = \frac{1}{b-a}\int_a^b f(x)\,dx \end{split} \end{equation*}

provided the limit exists.

We summarize this result in the following definition.

###### Definition3.1. Average Value of a Function.

If $f$ is an integrable function on the closed interval $[a,b]\text{,}$ then the average value $f_{avg}$ of $f$ on $[a,b]$ is

\begin{equation*} f_{avg} = \frac{1}{b-a} \int_a^b f(x)\,dx \end{equation*}

provided the definite integral exists.

Interactive Demonstration. Use the sliders below to investigate how changing $b = a + \Delta x$ changes the average value of the function. The area under the curve on the interval $[a,b]$ is shaded in blue.

###### Example3.2. Average Value Visualized.

Suppose a function is defined by $f(x)=16x^2+5\text{.}$

1. What is the average value of $f$ between $x=1$ and $x=3\text{?}$

2. Interpret your result from part (a) geometrically.

Solution
1. By definition of the average value of a function, we have

\begin{equation*} \begin{split} f_{avg} \amp = \frac{1}{3-1} \int_1^3 \left(16x^2 + 5\right)\,dx \\[1ex] \amp = \frac{1}{2} \left[\frac{16x^3}{3} + 5x\right]_1^3 = \frac{1}{2} \left[\left(16 \cdot 3^2 + 15\right) - \left(\frac{16 \cdot 1^3}{3} + 5\right)\right]\\[1ex] \amp =\frac{1}{2} \left(\frac{446}{3}\right) = \frac{223}{3} \end{split} \end{equation*}
2. We can interpret the same problem geometrically by asking the question: What is the average height of $f(x)=16x^2+5$ on the interval $[1,3]\text{?}$ The area $A$ under $f(x)=16x^2+5$ on the interval $[1,3]$ is given by

\begin{equation*} A = \int_1^3 \left(16x^2+5\right)\,dx = \frac{446}{3}\text{.} \end{equation*}

The area under the graph of $y=f_{avg}=\frac{223}{3}$ over the same interval $[1,3]$ is simply the area of a rectangle that is 2 units wide and $223/3$ units high with area $446/3$ units squared. So the average height of a function is is the height of the horizontal line that produces the same area over the given interval as shown below. ### Subsection3.1.2Area of Symmetric Functions

While we do not often work with even and odd functions, it is nonetheless useful to know the following facts about the area of the region under the curves of these functions on a symmetric interval $[-a,a]\text{.}$ Figure 3.(a) illustrates the area of the region under the curve of an even function $f$ on the interval $[-a,a]\text{,}$ and we are readily convinced that the net area shown is twice that of the net area of $f$ on the interval $[0,a]\text{.}$ Similarly, Figure 3.(b) illustrates the area of the region under the curve of an odd function $f$ on the interval $[-a,a]\text{,}$ and we readily believe that the net area is zero. (a) $f$ even function.

The proofs are readily accessible, so we only show the proof for item 1 of the theorem as the proof for item 2 is similar.

\begin{equation*} \begin{array}{lll} \ds\int_{-a}^a f(x)\,dx \amp = \ds{\int_{-a}^0 f(x)\,dx + \int_0^a f(x)\,dx} \amp \text{ sum rule } \\[2.5ex] \amp = \ds-\int_0^{-a} f(x)\,dx + \ds\int_0^a f(x)\,dx \amp \text{ order of limits rule } \\[2.5ex] \amp = \ds\int_0^a f(-u)\,du + \ds\int_0^a f(x)\,dx \amp u=-x,\ du = -dx,\ u(0)=0, \ u(-a)=a\\[2.5ex] \amp = \ds{\int_0^a f(u)\,du + \int_0^a f(x)\,dx} \amp f \text{ even } \\[2.5ex] \amp = \ds{2\int_0^a f(x)\,dx} \amp \text{ change dummy variables in first integral } \end{array} \end{equation*}

### Subsection3.1.3Area Between Curves: Integration w.r.t. $x$

We have seen how integration can be used to find an area between a curve and the $x$-axis. With very little change we can find some areas between curves; indeed, the area between a curve and the $x$-axis may be interpreted as the area between the curve and a second “curve” with equation $y=0\text{.}$

Suppose we would like to find the area below $\ds f(x)= -x^2+4x+3$ and above $\ds g(x)=-x^3+7x^2-10x+5$ over the interval $1\le x\le2\text{.}$ Figure 3.(a) depicts the area of the region contained between the curves of $f$ and $g$ as well as the lines $x=1$ and $x=2\text{.}$ We can see from this figure that the curve $f$ is indeed above the curve $g$ on the interval $[1,2]\text{.}$

We can approximate the area between two curves by dividing the area into thin sections and approximating the area of each section by a rectangle, as indicated in Figure 3.(b). The area of a typical rectangle is $\Delta x(f(x_i)-g(x_i))\text{,}$ so the total area is approximately

\begin{equation*} \sum_{i=0}^{n-1} (f(x_i)-g(x_i))\Delta x\text{.} \end{equation*}

This is exactly the sort of sum that turns into an integral in the limit, namely the integral

\begin{equation*} \int_1^2 f(x)-g(x)\,dx\text{.} \end{equation*}

Then

\begin{equation*} \begin{split} \int_1^2 f(x)-g(x)\,dx \amp =\int_1^2 (-x^2+4x+3)-(-x^3+7x^2-10x+5)\,dx \\[1.5ex] \amp = \int_1^2 x^3-8x^2+14x-2 \,dx = \left[\frac{x^4}{4} - \frac{8x^3}{3} + 7x^2 - 2x\right]_1^2\\[1.5ex] \amp = \left(\frac{16}{4} - \frac{64}{3} + 28 - 4\right) - \left(\frac{1}{4}-\frac{8}{3}+7-2\right)= 23-\frac{56}{3} - \frac{1}{4} = \frac{49}{12}. \end{split} \end{equation*} (a) Area of region between $f\text{,}$ $g\text{,}$ $x=1$ and $x=2$

We can also approach this problem of finding the area of the region between the curves of $f$ and $g$ as well as the lines $x=1$ and $x=2$ by using our knowledge of the definite integrals of $f$ and $g$ separately. Figure 3.(a) depicts the area of the region between the curves $f$ and the $x$-axis as well as the lines $x=1$ and $x=2\text{,}$ which we know as the definite integral

\begin{equation*} \int_1^2 f(x)\,dx\text{.} \end{equation*}

Similarly, Figure 3.(b) depicts the area of the region between the curves $g$ and the $x$-axis as well as the lines $x=1$ and $x=2\text{,}$ which we know as the definite integral

\begin{equation*} \int_1^2 g(x)\,dx\text{.} \end{equation*}

From Figure 3.4, we can readily determine that the area of the region we want must be the area of the region under the graph of $f$ minus the area of the region under the graph of $g$ as shown in Figure 3.(c). Once again, we have that

\begin{equation*} \begin{split} \int_1^2 f(x)-g(x) \,dx \amp = \int_1^2 \left(-x^2+4x+3\right)-\left(-x^3+7x^2-10x+5\right)\,dx= \frac{49}{12}. \end{split} \end{equation*} (a) $\int_1^2f(x)\ dx$

In summary, this procedure can informally be thought of as follows.

###### Area Between Two Curves: Integration w.r.t. $x$.
\begin{equation*} Area=\int_a^b (\mbox{top curve} ) - (\mbox{bottom curve} )\,dx,\qquad a\leq x\leq b\text{.} \end{equation*} This leads us to the following theorem for the area of the region between two curves on a closed interval.

Note: If we do not know which of the two functions $y=f(x)$ and $y=g(x)$ is larger on a given interval $[a,b]\text{,}$ we can still ensure that we calculate the area of the region between the curves of $f$ and $g$ and the lines $x=a$ and $x=b$ by placing absolute values around the integrand:
\begin{equation*} A=\ds\int_a^b \lvert f(x)-g(x) \rvert\,dx\text{.} \end{equation*}
###### Example3.5. Area Between Curves.

Find the area between $\ds f(x)= -x^2+4x$ and $\ds g(x)=x^2-6x+5\text{.}$

Solution

Here we are not given a specific interval, so it must be the case that there is a “natural” region involved, which we see from the graph shown below. Since the curves are both parabolas, the only reasonable interpretation is the region between the two intersection points.

We now find the intersection points by letting $f(x)=g(x)\text{:}$

\begin{equation*} \begin{gathered} -x^2+4x = x^2-6x + 5\\ -2x^2 +10x - 5 = 0 \\ \implies x = \frac{5}{2} \pm \frac{\sqrt{15}}{2}. \end{gathered} \end{equation*}

If we let

\begin{equation*} a=\frac{5-\sqrt{15}}{2} \text{ and } b=\frac{5+\sqrt{15}}{2}\text{,} \end{equation*}

the total area is

\begin{align*} \int_a^b \left(-x^2+4x\right)-(x^2-6x+5)\,dx \amp = \int_a^b -2x^2+10x-5\,dx\\ \amp = \left.-{2x^3\over3}+5x^2-5x\right|_a^b\\ \amp = 5\sqrt{15}\text{.} \end{align*}

after a bit of simplification. A general guideline to compute the area between two curves follows.

###### Guideline for Finding Area Between Two Curves: Integration w.r.t. $x$.

Given two continuous functions $y=f(x)$ and $y=g(x)$ on a closed interval $[a,b]\text{,}$ we can use the following steps to calculate the area of the region between the curves of $f$ and $g$ and the lines $x=a$ and $x=b\text{.}$

1. Find all intersection points of $f$ and $g$ in $[a,b]\text{.}$ Suppose there are $n$ of them called

\begin{equation*} i_1,i_2,\dots,i_n\text{.} \end{equation*}
2. Use these intersection points to create subintervals of $[a,b]\text{:}$

\begin{equation*} [a,b] = [a,i_1]\cup[i_1,i_2]\cup\dots\cup[i_n,b]\text{.} \end{equation*}
3. Draw a sketch of the two curves on $[a,b]$ and indicate all intersection points, similar to the figure shown below.

4. Based on the sketch, determine top curve and the bottom curve in each subinterval.

5. For each subinterval, write the area of the region like so

\begin{equation*} \int_{\text{ left bound } }^{\text{ right bound } } \left[\left(\text{ top curve } \right)-\left(\text{ bottom curve } \right)\right]\,dx \end{equation*}
6. The area of the region between the curves $f$ and $g$ and the lines $x=a$ and $x=b$ is the sum of all intervals from step 5, similar to the area calculated in Figure 3.5. Figure 3.5. The area of the region between the curves $f$ and $g$ and the lines $x=a$ and $x=b\text{.}$
###### Example3.6. Area Between Two Curves.

Determine the area enclosed by $y=x^2\text{,}$ $y=\sqrt x\text{,}$ $x=0$ and $x=2\text{.}$

Solution

The points of intersection of $y=x^2$ and $y=\sqrt x$ are

\begin{equation*} x^2=\sqrt x \implies x^4=x \implies x^4-x=0 \implies x(x^3-1)=0\text{.} \end{equation*}

Thus, either $x=0$ or $x=1\text{.}$ Sketching the curves gives: The area we want to compute is the shaded region. Since the top curve changes at $x=1\text{,}$ we need to use the formula twice. For $A_1$ we have $a=0\text{,}$ $b=1\text{,}$ the top curve is $y=\sqrt x$ and the bottom curve is $y=x^2\text{.}$ For $A_2$ we have $a=1\text{,}$ $b=2\text{,}$ the top curve is $y=x^2$ and the bottom curve is $y=\sqrt x\text{.}$

\begin{equation*} \text{Area} ~=~A_1+A_2 ~=~\int_0^1(\sqrt x-x^2)\,dx+\int_1^2(x^2-\sqrt x)\,dx \end{equation*}

For the first integral we have:

\begin{equation*} \int_0^1(\sqrt x-x^2)\,dx ~=~\left.\left(\frac{2}{3}x^{3/2}-\frac{1}{3}x^3\right)\right|_0^1 ~=~\frac{1}{3} \end{equation*}

Therefore, the second integral evaluates to

\begin{equation*} \int_1^2(x^2-\sqrt x)\,dx ~=~\left.\left(\frac{1}{3}x^3-\frac{2}{3}x^{3/2}\right)\right|_1^2 ~=~3-\frac{4\sqrt 2}{3} \end{equation*}

Thus,

\begin{equation*} \int_0^2 \vert x^2-\sqrt x\vert \,dx = A_1+A_2 = \frac{1}{3}+3-\frac{4\sqrt 2}{3} = \frac{10-4\sqrt 2}{3}\text{.} \end{equation*}
###### Example3.7. Area Between Sine and Cosine.

Determine the area enclosed by $y=\sin x$ and $y=\cos x$ on the interval $[0,~2\pi]\text{.}$

Solution

The curves $y=\sin x$ and $y=\cos x$ intersect when:

\begin{equation*} \sin x=\cos x\implies \tan x = 1\implies x=\frac{\pi}{4}+\pi k,~\mbox{$k$ an integer.} \end{equation*}

We have the following sketch: The area we want to compute is the shaded region. The top curve changes at $x=\pi/4$ and $x=5\pi/4\text{,}$ thus, we need to split the area up into three regions: from $0$ to $\pi/4\text{;}$ from $\pi/4$ to $5\pi/4\text{;}$ and from $5\pi/4$ to $2\pi\text{.}$

\begin{align*} \mbox{Area} \amp = \int_0^{\frac{\pi}{4}}(\cos x-\sin x)\,dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}(\sin x-\cos x)\,dx + \int_{\frac{5\pi}{4}}^{2\pi}(\cos x-\sin x)\,dx\\ \amp = \left(\sin x + \cos x\right)\bigg|_0^{\pi/4}+ \left(-\cos x - \sin x\right)\bigg|_{\pi/4}^{5\pi/4} + \left(\sin x + \cos x\right)\bigg|_{5\pi/4}^{2\pi}\\ \amp = \left(\sqrt 2-1\right) + \left(\sqrt 2+\sqrt 2\right) + \left(1+\sqrt 2\right)\\ \amp = 4\sqrt 2 \end{align*}

### Subsection3.1.4Area Between Curves: Integration w.r.t. $y$

Sometimes the given curves are not functions of $x\text{.}$ Instead, the area of the region is between two functions of $y\text{,}$ namely $x=f(y)$ and $x=g(y)\text{,}$ and the lines $y=c$ and $y=d\text{,}$ which can be thought of informally as the following.

###### Area Between Two Curves: Integration w.r.t. $y$.
\begin{equation*} Area=\int_c^d (\mbox{right curve} ) - (\mbox{left curve} )\,dy,\qquad c\leq y\leq d\text{.} \end{equation*} This leads us to a similar theorem as for the case when there are functions of $x$ bounding a region.

Note:

1. As before, if we do not know which of two functions $x=f(y)$ and $x=g(y)$ is larger on a given interval $[c,d]\text{,}$ we can still ensure that we calculate the area of the region between the curves of $f$ and $g$ and the lines $y=c$ and $y=d$ by placing absolute values around the integrand:

\begin{equation*} \int_c^d\vert f(y)-g(y)\vert\,dy \end{equation*}
2. Likewise, the Guideline for Finding Area Between Two Curves: Integration w.r.t. $x$ can be adopted for this scenario as well, which we leave to the reader to do.

###### Example3.9. Area Between Two Curves.

Determine the area enclosed by $x=y^2$ and $x=8\text{.}$

Solution

Note that $x=y^2$ and $x=8$ intersect when:

\begin{equation*} y^2=8\implies y=\pm\sqrt 8\implies y=\pm 2\sqrt 2 \end{equation*}

Sketching the two curves gives: From the sketch $c=-2\sqrt 2\text{,}$ $d=2\sqrt 2\text{,}$ the right curve is $x=8$ and the left curve is $x=y^2\text{.}$

\begin{equation*} \mbox{Area} ~=~\int_c^d[\mbox{right} -\mbox{left} ]\,dy ~=~\int_{-2\sqrt 2}^{2\sqrt 2} (8-y^2)\,dy ~=~\left.\left(8y-\frac{1}{3}y^3\right)\right|_{-2\sqrt 2}^{2\sqrt 2} \end{equation*}
\begin{equation*} = \left[8(2\sqrt 2)-\frac{1}{3}(2\sqrt 2)^3\right] - \left[8(-2\sqrt 2)-\frac{1}{3}(-2\sqrt 2)^3\right] ~=~\frac{64\sqrt 2}{3} \end{equation*}

We could have also integrated w.r.t. $x$ by using the two functions $y=\sqrt{x}$ and $y=-\sqrt{x}\text{.}$ We leave it to the reader to follow up on this integration.

##### Exercises for Section 3.1.

Find the average height of $f(x)$ over the given interval(s).

1. $f(x) = \cos x$ on $[0,\pi/2]\text{,}$ $[-\pi/2,\pi/2]\text{,}$ and $[0,2\pi]$

$2/\pi\text{;}$ $2/\pi\text{;}$ $0$
Solution

The average value of $f(x) = \cos(x)$ over $[0, \pi/2]$ is

\begin{equation*} f_{avg} = \frac{2}{\pi} \int_0^{\pi/2} \cos x\,dx = \frac{2}{\pi} \sin x \bigg\vert_0^{\pi/2} = \frac{2}{\pi}\text{.} \end{equation*}

The average value of $f(x) = \cos(x)$ over $[-\pi/2, \pi/2]$ is

\begin{equation*} f_{avg} = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos x\,dx = \frac{2}{\pi} \sin x \bigg\vert_{-\pi/2}^{\pi/2} = \frac{4}{\pi}\text{.} \end{equation*}

The average value of $f(x) = \cos(x)$ over $[0,2\pi]$ is

\begin{equation*} f_{avg} = \frac{2}{\pi} \int_{0}^{2\pi} \cos x\,dx = \frac{2}{\pi} \sin x \bigg\vert_{0}^{2\pi} = 0\text{.} \end{equation*}
2. $f(x)= x^2$ on $[-2,2]$

$4/3$
Solution

The average value of $f(x)=x^2$ over $[-2,2]$ is

\begin{equation*} f_{avg} = \frac{1}{2-(-2)} \int_{-2}^2 x^2\,dx = \frac{1}{4} \frac{x^3}{3}\bigg\vert_{-2}^2 = \frac{4}{3}\text{.} \end{equation*}
3. $f(x)=\dfrac{1}{x^2}$ on $[1,a]$

$1/a$
Solution

The average value of $f(x)=\frac{1}{x^2}$ on the interval $[1,a]$ (assuming $a > 1$), is

\begin{equation*} f_{avg} = \frac{1}{a-1} \int_1^a \frac{1}{x^2}\,dx = \frac{1}{a-1}\frac{-1}{x}\bigg\vert_1^a = \frac{1}{a-1}\left(\frac{-1}{a}+1\right) = \frac{1}{a}\text{.} \end{equation*}
4. $f(x)=\sqrt{1-x^2}$ on $[-1,1]$

$\pi/4$
Solution

The average value of $f(x) = \sqrt{1-x^2}$ on $[-1,1]$ is

\begin{equation*} f_{avg} = \frac{1}{2}\int_{-1}^1 \sqrt{1-x^2}\,dx = \frac{1}{4} \left[\sqrt(1 - x^2) x + \sin^{-1}(x)\right]_{-1}^1 = \frac{\pi}{4}\text{.} \end{equation*}

Find the area bounded by the curves by integrating with respect to $x\text{.}$

1. $\ds y=x^4-x^2\text{,}$ $\ds y=x^2$ (right of the $y$-axis)

$\ds 16\sqrt2/15$
Solution

Let $f(x) = x^4-x^2$ and $g(x)=x^2\text{.}$ We begin by plotting the two curves: The two curves intersect at $x=0$ and $x=\pm \sqrt{2}\text{.}$ The area $A$ between the curves (in green) is bounded above by $g(x)$ and below by $f(x)\text{,}$ and we notice that $A$ is symmetric about $x=0\text{.}$ Therefore, we calculate

\begin{equation*} A = 2\int_0^{\sqrt{2}} \left[g(x) - f(x)\right]\,dx = 2\int_0^{\sqrt{2}}\left[2x^2-x^4\right]\,dx = 2\left[\frac{2}{3}x^3-\frac{1}{5}x^5\right]_0^{\sqrt{2}} = \frac{16\sqrt{2}}{15}\text{.} \end{equation*}
2. $y=\cos(\pi x/2)\text{,}$ $\ds y=1- x^2$ (in first quadrant)

$2/3-2/\pi$
Solution

Let $f(x) = \cos(\pi x/2)$ and $g(x) = 1-x^2\text{.}$ We begin by plotting the two curves in the first quadrant: The two curves intersect at $(1,0)$ and $(0,1)\text{,}$ and the area $A$ (in green) is bounded above by $y=g(x)$ and bounded below by $y=f(x)\text{.}$ Therefore:

\begin{equation*} \begin{split} A \amp = \int_0^1 \left[g(x) - f(x)\right]\,dx \\ \amp = \int_0^1 \left[1 - x^2-\cos(\pi x/2)\right]\,dx \\ \amp = \left[-\frac{x^3}{3}+x - \frac{2}{\pi} \sin(\pi x/2)\right]_0^1 \\ \amp =\frac{2}{3}-\frac{2}{\pi}\\ \amp \approx 0.03005. \end{split} \end{equation*}
3. $y=\sin(\pi x/3)\text{,}$ $y=x$ (in first quadrant)

$\ds 3/\pi - 3\sqrt3/(2\pi)-1/8$
Solution

Let $f(x) = \sin(\pi x/3)$ and $g(x) = x\text{.}$ We begin by plotting the two curves in the first quadrant: The two curves intersect at $x=0$ and at $x=1/2\text{.}$ From the graph above, we see that $A$ is bounded above by $y=f(x)$ and bounded below by $y=g(x)\text{.}$ Therefore,

\begin{equation*} \begin{split} A \amp = \int_0^{1/2} \left[f(x)-g(x)\right]\,dx\\ \amp = \int_0^{1/2} \left[\sin(\pi x/3) - x\right]\,dx \\ \amp = \left[-\frac{3}{\pi} \cos(\pi x/3) -\frac{x^2}{2}\right]_0^{1/2}\\ \amp = \frac{3}{\pi} - \frac{2\sqrt{3}}{2\pi} - \frac{1}{8}\\ \amp \approx 0.0029363 \end{split} \end{equation*}
4. $\ds y=\sqrt{x}\text{,}$ $\ds y=x^2$

$1/3$
Solution

Let $f(x) = \sqrt{x}$ and $g(x) = x^2\text{.}$ We begin by plotting the two curves (for $x \geq 0$): The two curves intersect at $(0,0)$ and $(1,1)\text{.}$ The area $A$ (in green) is bounded above by the curve $y=f(x)$ and below by the curve $y=g(x)\text{.}$ Therefore, we compute:

\begin{equation*} \begin{split} A \amp = \int_0^1 \left[f(x) - g(x)\right]\,dx\\ \amp = \int_0^1 \left[\sqrt(x) - x^2\right]\,dx\\ \amp = \left[\frac{2 x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1\\ \amp = \frac{2}{3} - \frac{1}{3}= \frac{1}{3}. \end{split} \end{equation*}
5. $\ds y=x^{3/2}\text{,}$ $\ds y=x^{2/3}$

$1/5$
Solution

Let $f(x) = x^{3/2}$ and $g(x) = x^{2/3}\text{.}$ We begin by plotting the two curves: The two curves intersect at $(0,0)$ and $(1,1)\text{.}$ The area $A$ (in green) is bounded above by the curve $y=g(x)$ and below by the curve $y=f(x)\text{.}$ Therefore, we compute:

\begin{equation*} \begin{split} A \amp = \int_0^1 \left[g(x) - f(x)\right]\,dx\\ \amp = \int_0^1 \left[x^{2/3} - x^{3/2}\right]\,dx\\ \amp = \left[\frac{3x^{5/3}}{5} - \frac{2x^{5/2}}{5}\right]_0^1\\ \amp = \frac{3}{5} - \frac{2}{5} = \frac{1}{5} \end{split} \end{equation*}
6. $\ds y=x^2-2x\text{,}$ $y=x-2$

$1/6$
Solution

Let $f(x) = x^2-2x\text{,}$ $g(x) = x-2\text{.}$ First, we notice that

\begin{equation*} x^2-2x = x-2 \implies x =1, \ x=2\text{.} \end{equation*}

Now plot the two curves: So the area $A$ is bounded above by the curve $y=g(x)$ and below by the cruve $y=f(x)$ between $x=1$ and $x=2\text{.}$ We now compute:

\begin{equation*} \begin{split} A\amp = \int_1^2 \left[g(x)-f(x)\right]\,dx \\ \amp = \int_1^2 \left[(x-2)-(x^2-2x)\right]\,dx \\ \amp = \int_1^2 \left[-x^2+3x-2 \right]\,dx \\ \amp = \left[ -\frac{x^3}{3} + \frac{3 x^2}{2}- 2 x \right]_1^2\\ \amp = \frac{1}{6}. \end{split} \end{equation*}

Find the area bounded by the curves by integrating with respect to $y\text{.}$

1. $\ds x=y^3$ and $\ds x=y^2$

$1/12$
Solution

Let $f(y) = y^3$ and $g(y) = y^2\text{.}$ We first notice that

\begin{equation*} y^3=y^2 \implies y = 0, y = 1\text{.} \end{equation*}

Now plot the two curves: Therefore, the area $A$ is bounded to the left by the curve $x=f(y)$ and to the right by the curve $x=g(y)$ for $0 \leq y \leq 1\text{.}$ We now integrate with respect to $y\text{:}$

\begin{equation*} \begin{split} A \amp = \int_0^1 \left[g(y) - f(y)\right]\,dy \\ \amp = \int_0^1 \left[y^2 - y^3 \right]\,dy \\ \amp = \left[\frac{y^3}{3} - \frac{y^4}{4}\right]_0^1\\ \amp = \frac{1}{3}-\frac{1}{4} = \frac{1}{12}. \end{split} \end{equation*}
2. $x=0$ and $\ds x=25-y^2$

$500/3$
Solution

Let $f(y) = 0$ and $g(y) = 25-y^2\text{.}$ We first find the intersection points:

\begin{equation*} 25-y^2 = 0 \implies y = \pm 5\text{.} \end{equation*}

Now plot the curves: So for $-5 \leq y \leq 5\text{,}$ the area $A$ is bounded to the left by the curve $x=f(y)\text{,}$ and to the right by the curve $x=g(y)\text{.}$ We now integrate with respect to $y\text{:}$

\begin{equation*} \begin{split} A \amp = \int_{-5}^5 \left[g(y) - f(y)\right]\,dy\\ \amp = \int_{-5}^5 \left[25-y^2\right]\,dy\\ \amp = \left[25y - \frac{y^3}{3}\right]_{-5}^5\\ \amp = \frac{500}{3} \approx 166.67. \end{split} \end{equation*}

Find the area bounded by the curves.

1. $\ds x=1-y^2$ and $y=-x-1$

$9/2$
Solution

We first rewrite $y=-x-1$ as $x=-y-1\text{.}$ Now let $f(y) = 1-y^2$ and $g(y) =-y-1\text{.}$ The intersection points of these two curves is

\begin{equation*} 1-y^2 = -y-1 \implies y = -1,\ y=2\text{.} \end{equation*}

We now plot the two curves for $-1\leq y \leq 2\text{:}$ The area $A$ is bounded to the right by the curve $x=f(y)$ and to the left by the curve $x=g(y)$ on the entire interval $-1 \leq y \leq 2\text{.}$ We can now integrate with respect to $y\text{:}$

\begin{equation*} \begin{split} A \amp = \int_{-1}^2 \left[f(y) -g(y)\right]\,dy \\ \amp = \int_{-1}^2 \left[2-y^2 + y\right]\,dy\\ \amp = \left[2y - \frac{y^3}{3} + \frac{y^2}{2} \right]_{-1}^2\\ \amp = \frac{9}{2} = 4.5 \end{split} \end{equation*}
2. $\ds x=3y-y^2$ and $x+y=3$

$4/3$
Solution

Let $f(y)=3y-y^2$ and $g(y)=3-y\text{.}$ Then we wish to find the area between the curves $x=f(y)$ and $x=g(y)\text{:}$ The curves intersect at the points $(0,3)$ and $(2,1)\text{,}$ and we notice that the desired area $A$ (in green) is bounded to the left by $g(y)$ and to the right by $f(y)\text{.}$ We therefore integrate with respect to $y\text{:}$

\begin{equation*} A = \int_1^3 \left[3y-y^2 - (3-y)\right]\,dy = \int_1^3 (4y-y^2 - 3)\,dy =\left[2y^2-\frac{1}{3}y^3-3y\right]_1^3 = \frac{4}{3}\text{.} \end{equation*}
3. $\ds y=\sqrt x$ and $\ds y=\sqrt{x+1}\text{,}$ $0\le x\le 4$

$\ds 10\sqrt{5}/3-6$
Solution

Let $f(x) = \sqrt{x}$ and $g(x) = \sqrt{x+1}\text{.}$ The area between the two curves $f$ and $g\text{,}$ as well as the lines $x=0$ and $x=4$ is the shaded area below: $A$ is bounded below by the curve $y=f(x)$ and is bounded above by the curve $y=g(x)$ for all $x \in [0,4]\text{.}$ Therefore, we integrate with respect to $x\text{:}$

\begin{equation*} \begin{split} A \amp = \int_0^4 \left[g(x) - f(x)\right]\,dx\\ \amp = \int_0^4 \left[\sqrt{x+1} - \sqrt{x}\right]\,dx\\ \amp = \left[\frac{2(x+1)^{2/3}}{3} - \frac{2x^{3/2}}{3}\right]_0^4\\ \amp = \frac{10\sqrt{5}}{3} - 6 \approx 1.4536. \end{split} \end{equation*}
4. $y=\sin x\cos x$ and $y=\sin x\text{,}$ $0\le x\le \pi$

$2$
Solution

Let $f(x) = \sin x \cos x$ and $g(x) = \sin x\text{.}$ We first rewrite $f(x) = \frac{1}{2} \sin(2x)\text{.}$ Therefore, the area between the curves $f$ and $g$ for $x \in [0,\pi]$ is the shaded area below: The curves intersect at $(0,0)$ and at $(\pi,0)\text{.}$ The desired area $A$ is bounded above by $g(x)$ and below by $f(x)$ for all $x \in [0,\pi]\text{.}$ We therefore integrate with respect to $x\text{:}$

\begin{equation*} A = \int_0^\pi \left(\sin(x)-\frac{1}{2}\sin(2x)\right)\,dx = \left[\frac{1}{4}\cos(2x)-\cos(x)\right]_0^\pi = \frac{1}{4} + 1 - \frac{1}{4} + 1 = 2\text{.} \end{equation*}

Find the area bounded by the given curves. 6.75
Solution 1

The area $A$ is bounded above by the curve $y=0$ and above by the curve $y=x^3-3x^2$ for all $x \in [0,3]\text{.}$ We therefore integrate with respect to $x\text{:}$

\begin{equation*} \begin{split} A \amp = \int_0^3 \left[0 - x^3+3x^2\right]\,dx \\ \amp = \left[-\frac{x^4}{4} + x^3\right]_0^3\\ \amp = -\frac{3^4}{4} + 3^3 = \frac{27}{4} = 6.75. \end{split} \end{equation*} 12.15
Solution 2

For $0 \leq x \leq 3\text{,}$ the area $A$ is bounded above by $y=0$ and below by $y=x^4-3x^3\text{.}$ We therefore integrate with respect to $x\text{:}$

\begin{equation*} \begin{split} A\amp = \int_0^3 \left[0-x^4+3x^3\right]\,dx\\ \amp = \left[-\frac{x^5}{5} + \frac{3x^4}{4}\right]_0^3\\ \amp =\frac{243}{20} = 12.15 \end{split} \end{equation*} 16/3
Solution 3

We first notice the odd symmetry in this problem. On $[0,2]\text{,}$ the area is bounded above by $y=x\sqrt{4-x^2}$ and below by $y=0\text{.}$ Therefore, the total area can be expressed as

\begin{equation*} A = 2\int_0^2 x\sqrt{4-x^2}\,dx \end{equation*}

Let $u=4-x^2\text{,}$ with $du = -2x\,dx\text{.}$ Then when $x=0\text{,}$ $u=4$ and when $x=2\text{,}$ $u=0\text{.}$ Therefore:

\begin{equation*} \begin{split} A \amp = 2 \int_4^0 \sqrt{u} \frac{du}{-2}\\ \amp = \int_0^4 \sqrt{u}\,du\\ \amp = \frac{2u^{2/3}}{3} \bigg\vert_0^4\\ \amp =\frac{16}{3} \approx 5.33 \end{split} \end{equation*} $3\ln(2)$
Solution 4

First notice that the curve $y=\displaystyle{\frac{3x}{x^2+1}}$ is an odd function. Therefore, the desired area is given by

\begin{equation*} A = \int_{-1}^0 \left[0-\frac{3x}{x^2+1}\right]\,dx + \int_0^1 \left[\frac{3x}{x^2+1}-0\right]\,dx = 2\int_0^1 \frac{3x}{x^2+1}\,dx\text{.} \end{equation*}

We carry out the integration by making the substitution:

\begin{equation*} u = x^2+1, \ du = 2x \,dx\text{.} \end{equation*}

Then,

\begin{equation*} \begin{split} \int_0^1 \frac{3x}{x^2+1}\,dx \amp = \frac{3}{2}\int_1^2 \frac{du}{u} \\[1ex] \amp = \frac{3}{2} \ln |u| \bigg\vert_1^2 \\[1ex] \amp = \frac{3}{2} \ln 2 \end{split} \end{equation*}

The area is thus

\begin{equation*} A = 2\int_0^1 \frac{3x}{x^2+1}\,dx = 2 \left(\frac{3}{2} \ln 2\right) = 3\ln 2\text{.} \end{equation*} 4.5
Solution 5

The area is bounded above by the curve $y=0$ and below by the curve $y=\frac{x}{3} - \sqrt{x}$ for $x \in [0,9]\text{.}$ We therefore integrate with respect to $x\text{:}$

\begin{equation*} \begin{split} A \amp = \int_0^9 \left[0-\frac{x}{3} + \sqrt{x}\right]\,dx\\ \amp = \left[-\frac{x^2}{6} + \frac{2 x^{3/2}}{3}\right]_0^9\\ \amp = \frac{9}{2} = 4.5 \end{split} \end{equation*} 1/3
Solution 6

For $x\in[0,1]\text{,}$ the area is bounded above by the curve $y=\sqrt{x/4}$ and below by the curve $y=-1/2+x\text{.}$ We therefore integrate with respect to $x\text{:}$

\begin{equation*} \begin{split} A \amp = \int_0^1 \left[\sqrt{\frac{x}{4}} + \frac{1}{2} - x\right]\,dx \\ \amp = \left[\frac{x^{3/2}}{3} - \frac{x^2}{2} + \frac{x}{2}\right]_0^1\\ \amp = \frac{1}{3} - \frac{1}{2} + \frac{1}{2} = \frac{1}{3}. \end{split} \end{equation*}

An object moves so that its velocity at time $t$ is $v(t)=-9.8t+20$ m/s. Describe the motion of the object between $t=0$ and $t=5\text{,}$ find the total distance travelled by the object during that time, and find the net distance travelled. Solution

It rises until $t=100/49\text{,}$ then falls. The position of the object at time $t$ is $\ds s(t)=-4.9t^2+20t+k\text{.}$ The net distance travelled is $-45/2\text{,}$ that is, it ends up $45/2$ meters below where it started. The total distance travelled is $6205/98$ meters.

An object moves so that its velocity at time $t$ is $v(t)=\sin t\text{.}$ Set up and evaluate a single definite integral to compute the net distance travelled between $t=0$ and $t=2\pi\text{.}$

$\ds\int_0^{2\pi}\sin t\,dt=0$
Solution

We make a sketch of the velocity function below: By symmetry of the sine function, we see that the net distance travelled must be

\begin{equation*} \int_0^{2\pi}\sin t\,dt=0\text{.} \end{equation*}

An object moves so that its velocity at time $t$ is $v(t)=1+2\sin t$ m/s. Find the net distance travelled by the object between $t=0$ and $t=2\pi\text{,}$ and find the total distance travelled during the same period.

net: $2\pi\text{,}$ total: $\ds 2\pi/3+4\sqrt3$
Solution

We make a sketch of the velocity function below: Therefore, the net distance travelled is

\begin{equation*} \int_0^{2\pi} \left(1+2\sin t\right)\,dt = 2\pi\text{.} \end{equation*}

And the total distance $D$ travelled is the total area under the velocity curve. Since the velocity is negative for $t\in \left(\frac{7\pi}{6},\frac{11\pi}{6}\right)\text{,}$ we split the integral into three pieces:

\begin{equation*} \begin{split} D \amp = \int_0^{7\pi/6} (1+2\sin t)\,dt - \int_{7\pi/6}^{11\pi/6} (1+2\sin t)\,dt + \int_{11\pi/6}^{2\pi} (1+2\sin t)\,dt \\ \amp = \left[t-2\cos t\right]_0^{7\pi/6} - \left[t-2\cos t\right]_{7\pi /6}^{11\pi/6} + \left[t-2\cos t\right]_{11\pi/6}^{2\pi}\\ \amp = \left(2+\sqrt{3} + \frac{7\pi}{6}\right) - \left(\frac{2\pi}{3} - 2\sqrt{3}\right) + \left(-2 + \sqrt{3} + \frac{\pi}{6}\right) \\ \amp = 4\sqrt{3} + \frac{2\pi}{3}. \end{split} \end{equation*}

Consider the function $f(x)=(x+2)(x+1)(x-1)(x-2)$ on $[-2,2]\text{.}$ Find the total area between the curve and the $x$-axis (measuring all area as positive).

$8$
Solution

We first sketch the function $f(x)$ on $[-2,2]\text{:}$ We notice that the area is bounded below by the curve $y=f(x)$ on $(-2,-1)\cup(1,2)\text{,}$ and is bounded above by the curve on $(-1,1)\text{.}$ Therefore, the total area will be given by:

\begin{equation*} A = \int_{-2}^{-1} \left[0-f(x)\right]\,dx + \int_{-1}^{1} \left[f(x)-0\right]\,dx + \int_1^2 \left[0-f(x)\right]\,dx\text{.} \end{equation*}

Furthermore, we note that the area is symmetric about the $y$-axis. Therefore, we can simplify:

\begin{equation*} A = 2\left[\int_0^1 f(x)\,dx - \int_1^2 f(x)\,dx\right]\text{.} \end{equation*}

First, we solve the corresponding indefinite integral by rewriting the integrand in expanded form:

\begin{equation*} \int f(x)\,dx = \int (x^4-5x^2+4) \,dx = \frac{x^5}{5} - \frac{5x^3}{3} + 4x + C\text{.} \end{equation*}

We now evaluate the definite integrals:

\begin{equation*} \begin{split} A \amp = 2\left[\int_{0}^1 f(x)\,dx - \int_1^2 f(x)\,dx\right] \\ \amp = 2\left[\frac{x^5}{5} - \frac{5x^3}{3} + 4x\right]_{0}^1 - 2\left[\frac{x^5}{5} - \frac{5x^3}{3} + 4x\right]_1^2\\ \amp = 2 \frac{38}{15} + 2 \frac{22}{15} = 8. \end{split} \end{equation*}

Consider the function $\ds f(x)=x^2-3x+2$ on $[0,4]\text{.}$ Find the total area between the curve and the $x$-axis (measuring all area as positive).

$17/3$
Solution

We first rewrite $f(x) = x^2-3x+2 = (x-1)(x-2)\text{.}$ From the plot above, we see that the area (in green) is bounded above by the curve $y=f(x)$ on $(0,1)\cup(2,4)\text{,}$ and below by the curve on $(1,2)\text{.}$ Therefore, the area is given by

\begin{equation*} A = \int_0^1 \left[f(x) -0\right]\,dx + \int_1^2 \left[0-f(x)\right]\,dx + \int_2^4 \left[f(x)-0\right]\,dx\text{.} \end{equation*}

We first find the corresponding indefinite integral:

\begin{equation*} \int f(x)\,dx = \int (x^2-3x+2)\,dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x + C\text{.} \end{equation*}

Using this result, we can compute the definite integrals:

\begin{equation*} \begin{split} A \amp = \int_0^1 f(x)\,dx - \int_1^2 f(x)\,dx + \int_2^4 f(x)\,dx\\ \amp = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_0^1 - \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_1^2 + \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_2^4\\ \amp = \frac{5}{6} - \left(-\frac{1}{6}\right) + \frac{28}{6} = \frac{17}{3}. \end{split} \end{equation*}

Evaluate the three integrals:

\begin{equation*} A=\int_0^3 (-x^2+9)\,dx\qquad B=\int_0^{4} (-x^2+9)\,dx\qquad C=\int_{4}^3 (-x^2+9)\,dx\text{,} \end{equation*}

and verify that $A=B-C\text{.}$ Draw a diagram which illustrates this relationship.

$A=18\text{,}$ $B=44/3\text{,}$ $C=10/3$
Solution

We first evaluate the corresponding indefinite integral:

\begin{equation*} \int (-x^2 + 9)\,dx = -\int x^2 \,dx + 9\int \,dx = -\frac{x^3}{3} + 9x + C\text{.} \end{equation*}

Therefore,

\begin{equation*} A = \int_0^3 (-x^2 + 9)\,dx = \left[-\frac{x^3}{3} + 9x\right]_0^3 = 18\text{.} \end{equation*}

Similarly, we have

\begin{equation*} B = \int_0^4 (-x^2+9)\,dx = \left[ -\frac{x^3}{3} + 9x\right]_0^4 = \frac{44}{3}\text{,} \end{equation*}

and

\begin{equation*} C = \int_3^4 (-x^2+9)\,dx = \left[ -\frac{x^3}{3} + 9x\right]_3^4 = -\frac{10}{3}\text{.} \end{equation*}

Hence,

\begin{equation*} B - C = \frac{44}{3} + \frac{10}{3} = \frac{54}{3} = 18 = A\text{.} \end{equation*}

We illustrate this relationship below: where $B$ is the net area—in other words, $B=A+C\text{.}$