Area of a Region Between Two Curves
Area of a Region Between Two Curves
If $f$ and $g$ are continuous on $[a, b]$ and $g(x) \leq f(x)$ for all $x$ on $[a, b]$, then the area of the region bounded by the graphs of $f$ and $g$ and the vertical lines $x=a$ and $x=b$ is
$
A=\int_a^b[f(x)-g(x)] d x
$
In general, to determine the area between two curves, we can use
$
A=\int_{x_1}^{x_2}[(\text { top curve) }-(\text { bottom curve })] d x
$
if the curves are defined by functions of $x$, and
$
A=\int_{y_1}^{y_2}[(\text { right curve })-(\text { left curve })] d y,
$
if the curves are defined by functions of $y$.
- Example 1
Find the area of the region in the first quadrant enclosed by the graphs of $f(x)=2 \cos x, g(x)=x$ and the $y$-axis.
▶️Answer/Explanation
Solution
The limits of integration are $a=0$ and $b=1.03$.
Use a graphing calculator to find the point of intersection and sketch the two curves. The point of intersection in the first quadrant is $(1.03,1.03)$.
The area between the curves is
$
\begin{aligned}
A & =\int_a^b[f(x)-g(x)] d x \\
& =\int_0^{1.03}[2 \cos x-x] d x \\
& =\left[2 \sin x-\frac{x^2}{2}\right]_0^{1.03} \\
& =\left(2 \sin 1.03-\frac{1.03^2}{2}\right)-(2 \sin 0-0) \\
& =1.184
\end{aligned}
$
Example 2
- What is the area of the region enclosed by the graphs of $f(x)=x+2$ and $g(x)=x^3-4 x^2+6$ ?
(A) $\frac{193}{12}$
(B) $\frac{218}{12}$
(C) $\frac{253}{12}$
(D) $\frac{305}{12}$
▶️Answer/Explanation
Ans:C
Example 3
- What is the area of the region in the first quadrant, bounded by the curve $y=\sqrt[3]{x}$ and $y=x$ ?
(A) $\frac{1}{5}$
(B) $\frac{1}{4}$
(C) $\frac{1}{3}$
(D) $\frac{1}{2}$
▶️Answer/Explanation
Ans:B
Example5
The curve $y=f(x)$ and the line $y=-3$, shown in the figure above, intersect at the points $(0,-3)$, $(a,-3)$, and $(b,-3)$. The sum of area of the shaded region enclosed by the curve and the line is given by
(A) $\int_0^a[3-f(x)] d x+\int_a^b[-3+f(x)] d x$
(B) $\int_0^a[-3+f(x)] d x+\int_a^b[3-f(x)] d x$
(C) $\int_0^a[f(x)+3] d x+\int_a^b[-3-f(x)] d x$
(D) $\int_0^a[f(x)-3] d x+\int_a^b[3-f(x)] d x$
▶️Answer/Explanation
Ans:C
Volumes by Disk and Washers
The solid generated by rotating a plane region about an axis is called a solid of revolution.
The simplest such solid is a right circular cylinder or disk.
Volume of disk = (base area of disk)(width of disk)
$\Delta V=A(x) \Delta x$.
So the definition of volume gives us $V=\int_a^b A(x) d x$.
The Disk Method
Horizontal Axis of Revolution Vertical Axis of Revolution
$
V=\pi \int_a^b[R(x)]^2 d x \quad\quad \quad V=\pi \int_{\mathrm{c}}^{\mathrm{d}}[R(y)]^2 d y
$
The Washer Method
Horizontal Axis of Revolution Vertical Axis of Revolution
$V=\pi \int_a^b\left([R(x)]^2-[r(x)]^2\right) d x \quad V=\pi \int_c^d\left([R(y)]^2-[r(y)]^2\right) d y$
Example 2
- Find the volume of the solid generated by revolving the region bounded by the graphs of $y=\sqrt{x}, x=1$, and $x=4$ about the $x$-axis.
▶️Answer/Explanation
Solution
Sketch the graphs. The volume is
$
\begin{aligned}
V & =\pi \int_a^b[R(x)]^2 d x \\
& =\pi \int_1^4[\sqrt{x}]^2 d x=\pi \int_1^4 x d x \\
& =\pi\left[\frac{x^2}{2}\right]_1^4=\pi\left(\frac{16}{2}-\frac{1}{2}\right) \\
& =\frac{15 \pi}{2}
\end{aligned}
$
Example 3
- The region in the first quadrant bounded by the graph of $y=\sec x, x=\frac{\pi}{3}$, and the coordinate axes is rotated about the $x$-axis. What is the volume of the solid generated?
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{2}$
(C) $\sqrt{3} \pi$
(D) $3 \pi$
▶️Answer/Explanation
Ans:C
Example 4
- The region enclosed by the graphs of $y=e^{(x / 2)}$ and $y=(x-1)^2$ from $x=0$ to $x=1$ is rotated about the $x$-axis. What is the volume of the solid generated?
(A) $\frac{11}{4} \pi$
(B) $2(e-1) \pi$
(C) $\left(e-\frac{3}{2}\right) \pi$
(D) $\left(e-\frac{6}{5}\right) \pi$
▶️Answer/Explanation
Ans:D
Example 5
Let $R$ be the region between the graphs of $y=1+\sin (\pi x)$ and $y=x^3$ from $x=0$ to $x=1$. The volume of the solid obtained by revolving $R$ about the $x$-axis is given by
(A) $\pi \int_0^1\left[1+\sin (\pi x)-x^3\right] d x$
(B) $\pi \int_0^1\left[(1+\sin (\pi x))^2-x^6\right] d x$
(C) $\pi \int_0^1\left[1+\sin (\pi x)-x^3\right]^2 d x$
(D) $2 \pi \int_0^1\left[1+\sin (\pi x)-x^3\right] d x$
▶️Answer/Explanation
Ans:B
Example 6
- Find the volume of the solid formed by revolving the region bounded by the graphs of $y=x^{1 / 3}, y=x^2$,
(a) about the $x$-axis
(b) about the $y$-axis.
▶️Answer/Explanation
Solution
Sketch the graphs.
(a) The region runs from $x=0$ to $x=1$.
The volume is
$
\begin{aligned}
V & =\pi \int_a^b\left([R(x)]^2-[r(x)]^2\right) d x \\
& =\pi \int_0^1\left(\left[x^{1 / 3}\right]^2-\left[x^2\right]^2\right) d x \\
& =\pi \int_0^1\left(x^{2 / 3}-x^4\right) d x \\
& =\pi\left[\frac{3}{5} x^{5 / 3}-\frac{1}{5} x^5\right]_0^1 \quad y= \\
& =\frac{2 \pi}{5}
\end{aligned}
$
(b) The region runs from $y=0$ to $y=1$.
$
\begin{aligned}
V & =\pi \int_c^d\left([R(y)]^2-[r(y)]^2\right) d y \\
& =\pi \int_0^1\left([\sqrt{y}]^2-\left[y^3\right]^2\right) d y \\
& =\pi \int_0^1\left(y-y^6\right) d x \\
& =\pi\left[\frac{1}{2} y^2-\frac{1}{7} y^7\right]_0^1 \\
& =\frac{5 \pi}{14}
\end{aligned}
$
Volumes of Solids with Known Cross Sections
The volume of a solid of known cross-sectional area $A(x)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$.
1. For cross sections of area $A(x)$ taken perpendicular to the $x$-axis, the volume is
$
V=\int_a^b A(x) d x
$
2. For cross sections of area $A(y)$ taken perpendicular to the $y$-axis, the volume is
$
V=\int_c^d A(y) d y
$
Example 1
- The base of a solid is the region in the first quadrant enclosed by the graph of $y=-x^2+2 x$ and the $x$-axis. If every cross section of the solid perpendicular to the $x$-axis is an equilateral triangle, what is the volume of the solid?
▶️Answer/Explanation
Solution
Sketch the typical cross section.
The area of each cross section is
$
\begin{aligned}
& A(x)=\frac{1}{2}(\text { base })(\text { height })=\frac{1}{2}(y)\left(\frac{\sqrt{3}}{2} y\right)=\frac{\sqrt{3}}{4} y^2 \\
& =\frac{\sqrt{3}}{4}\left(-x^2+2 x\right)^2 \\
& =\frac{\sqrt{3}}{4}\left(x^4-4 x^3+4 x^2\right) \\
&
\end{aligned}
$
The equilateral triangles lie on the plane from $x=0$ to $x=2$. The volume is
$
\begin{aligned}
V & =\int_a^b A(x) d x=\int_0^2 \frac{\sqrt{3}}{4}\left(x^4-4 x^3+4 x^2\right) d x \\
& =\frac{\sqrt{3}}{4}\left[\frac{x^5}{5}-x^4+\frac{4}{3} x^3\right]_0^2=\frac{4 \sqrt{3}}{15}
\end{aligned}
$
Example 2
- The base of a solid is the region enclosed by the graph of $y=e^x$, the coordinate axes, and the line $x=1$. If the cross sections of the solid perpendicular to the $x$-axis are squares, what is the volume of the solid?
(A) $\frac{e^2}{4}$
(B) $\frac{e^2-1}{2}$
(C) $\frac{e^2+1}{2}$
(D) $e^2-\frac{1}{2}$
▶️Answer/Explanation
Ans:B
Example 3
- The base of a solid is the region enclosed by the graph of $y=\sqrt{x}$, the $x$-axis, and the line $x=2$. If each cross section perpendicular to the $x$-axis is an equilateral triangle, what is the volume of the solid?
(A) $\frac{\sqrt{3}}{8}$
(B) $\frac{\sqrt{3}}{6}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{3}}{2}$
▶️Answer/Explanation
Ans:D
Example 4
- The base of a solid is the region in the first quadrant bounded by the coordinate axes, and the line $2 x+3 y=6$. If the cross sections of the solid perpendicular to the $x$-axis are semicircles, what is the volume of the solid?
(A) $\frac{\pi}{2}$
(B) $\frac{3 \pi}{4}$
(C) $\pi$
(D) $\frac{3 \pi}{2}$
▶️Answer/Explanation
Ans:A
Example 5
- The base of a solid $S$ is the semicircular region enclosed by the graph of $y=\sqrt{9-x^2}$ and the $x$-axis. If the cross sections of $S$ perpendicular to the $x$-axis are semicircles, what is the volume of the solid?
(A) $\frac{20 \pi}{3}$
(B) $6 \pi$
(C) $\frac{9 \pi}{2}$
(D) $\frac{7 \pi}{2}$
▶️Answer/Explanation
Ans:C
Example 6
- The base of a solid is the region bounded by the graph of $y=\sqrt{x}$, the $x$-axis and the line $x=4$. If the cross sections of the solid perpendicular to the $y$-axis are squares, the volume of the solid is given by
(A) $\int_0^2\left(4-y^2\right)^2 d y$
(B) $\int_0^2(4-y)^2 d y$
(C) $\int_0^2\left[(2-y)^2\right]^2 d y$
(D) $\int_0^4\left[(2-y)^2\right]^2 d y$
▶️Answer/Explanation
Ans:A
The Total Change Theorem (Application of FTC)
The total change (accumulated change) in a quantity over a time period i the rate of change of the quantity.
$
\text { Total Change }=\int_a^b F^{\prime}(x) d x=F(b)-F(a)
$
This principle can be applied to all of the rates of change in natural and social sciences. The following are a few examples of these idea:
1. If $F^{\prime}(t)$ is the rate of growth of a population at time $t$, then
$
\int_a^b F^{\prime}(t) d t=F(b)-F(a)
$
is the increase in population between times $t=a$ and $t=b$.
2. If $F^{\prime}(t)$ is the rate of decomposition of a certain chemical, then
$
\int_a^b F^{\prime}(t) d t=F(b)-F(a)
$
is the total amount that has decomposed between times $t=a$ and $t=b$.
3. If $F^{\prime}(t)$ is the rate of consumption of a certain product, then
$
\int_a^b F^{\prime}(t) d t=F(b)-F(a)
$
is the total amount that has been consumed between times $t=a$ and $t=b$.
4. If $F^{\prime}(t)$ is the rate of change of temperature in a room, then
$
\int_a^b F^{\prime}(t) d t=F(b)-F(a)
$
is the total change in temperature between times $t=a$ and $t=b$.
Example 1
- For $0 \leq t \leq 60$, the rate of change of the number of mosquitoes at time $t$ days is modeled by $f(t)=6 \sqrt{t} \sin \left(\frac{t}{6}\right)$ mosquitoes per day. There are 1200 mosquitoes at time $t=0$.
(a) At time $t=20$, is the number of mosquitoes increasing or decreasing?
(b) According to the model, how many mosquitoes will be there at time $t=60$.
(c) To the nearest whole number, what is the maximum number of mosquitoes for $0 \leq t \leq 60 ?$
▶️Answer/Explanation
Solution
(a) Since $f(20)=6 \sqrt{20} \sin \left(\frac{20}{6}\right)=-5.113<0$, the number of mosquitoes is decreasing at $t=20$.
(b) The number of mosquitoes at time $t=60$ is
$
\begin{array}{ll}
1200+\int_a^b f(t) d t & \text { Initially there were } 1200 \text { mosquitoes. } \\
=1200+\int_0^{60} 6 \sqrt{t} \sin \left(\frac{t}{6}\right) d t & \int_0^{60} 6 \sqrt{t} \sin \left(\frac{t}{6}\right) d t \text { represents the increase in } \\
& \text { mosquitoe population between times } t=0 \text { and } t=60 . \\
=1200+282.272=1482.272 & \text { Use a graphing calculator to find the value of } \\
& \int_0^{60} 6 \sqrt{t} \sin \left(\frac{t}{6}\right) d t .
\end{array}
$
(c) $f(t)=0$ when $t=0, t=6 \pi, t=12 \pi, t=18 \pi$
The absolute maximum number of mosquitoes occurs at $t=6 \pi$ or $t=18 \pi$.
$
\begin{gathered}
1200+\int_0^{6 \pi} 6 \sqrt{t} \sin \left(\frac{t}{6}\right) d t=1200+214.751=1414.751 \\
1200+\int_0^{18 \pi} 6 \sqrt{t} \sin \left(\frac{t}{6}\right) d t=1200+326.710=1526.710
\end{gathered}
$
There are 1415 mosquitoes at $t=6 \pi$ and 1527 mosquitoes at $t=18 \pi$, so the maximum number of mosquitoes is 1527 to the nearest whole number.
Example 2
- Water leaks from a tank at a rate of $r(t)$ gallons per hour. The graph of $r$, for the time interval $0 \leq t \leq 10$ hours, is shown at the right. Use the midpoint Riemann sum with five intervals of equal length to estimate the total amount of water leaked out during the first 10 hours.
▶️Answer/Explanation
Solution
The five intervals are $[0,2],[2,4],[4,6],[6,8]$, and $[8,10]$. $75,70,60,45$, and 25 are the heights at each midpoint.
Total amount of water leaked out during the first 10 hours is $75 \cdot 2+70 \cdot 2+60 \cdot 2+45 \cdot 2+25 \cdot 2$ $=550$ gallons .
Example 3
Oil is pumped out from a tank at the rate of $\frac{20 e^{(-0.1 t)}}{1+e^{-t}}$ gallons per minute, where $t$ is measured in minutes. To the nearest gallon, how many gallons of oil are pumped out from a tank during the time interval $0 \leq t \leq 6$ ?
(A) 62
(B) 78
(C) 85
(D) 93
▶️Answer/Explanation
Ans:B
Example 4
Pollutant is released into a lake at the rate of $\frac{50 e^{-t / 2}}{\sqrt{t+1}}$ gallons per hour. To the nearest gallon, how many gallons of pollutant are released during the time interval $0 \leq t \leq 12$ ?
(A) 53
(B) 58
(C) 66
(D) 75
▶️Answer/Explanation
Ans:C
Example5
Oil is pumped into an oil tank at the rate of $S(t)$ gallons per hour during the time interval $0 \leq t \leq 8$ hours. During the same time interval, oil is removed from the tank at the rate of $R(t)$ gallons per hour. If the oil tank contained 200 gallons of oil at time $t=0$, which of the following expressions shows the amount of oil in the tank at time $t=6$ hours?
(A) $200+S(6)-R(6)$
(B) $200+S^{\prime}(6)-R^{\prime}(6)$
(C) $200+\int_0^6(S(t)-R(t)) d t$
(D) $200+\int_0^6\left(S^{\prime}(t)-R^{\prime}(t)\right) d t$
▶️Answer/Explanation
Ans:C
Motion of a Particle, Distance, and Displacement
If a particle moves along a straight line with position function $s(t)$, then its velocity is $v(t)=s^{\prime}(t)$, so
$
s(b)-s(a)=\int_a^b v(t) d t
$
is the change of the position, or displacement, of the particle during the time period from $t=a$ to $t=b$.
The average velocity and the average acceleration over the time interval from $t=a$ to $t=b$ is
Average velocity $=\frac{\text { displacement }}{\text { time }}=\frac{s(b)-s(a)}{b-a}=\frac{1}{b-a} \int_a^b v(t) d t$
Average acceleration $=\frac{v(b)-v(a)}{b-a}=\frac{1}{b-a} \int_a^b a(t) d t$
To find the total distance traveled we have to consider when the particle moves to the right, $v(t) \geq 0$, and when the particle moves to the left, $v(t) \leq 0$.
In both cases, the distance is computed by integrating $|v(t)|$, the speed of the particle. Therefore
Total distance traveled $=\int_a^b|v(t)| d t$
Average speed $=\frac{\text { distance traveled }}{\text { time }}=\frac{1}{b-a} \int_a^b|v(t)| d t$.
The acceleration of the object is $a(t)=v^{\prime}(t)$, so
$
v(b)-v(a)=\int_a^b a(t) d t
$
is the change in velocity from $t=a$ to $t=b$.
The figure at the right shows how both displacement and distance traveled can be interpreted in terms of areas under a velocity curve.
displacement
$
=\int_a^b v(t) d t=A_1-A_{\mathbf{2}}+A_3
$
total distance traveled
$
=\int_a^b|v(t)| d t=A_1+A_2+A_3
$
Example 1
- A particle moves along the $x$-axis so that its velocity at any time $t \geq 0$ is given by $v(t)=t^2-3 t-4$.
(a) In which direction (left or right) is the particle moving at time $t=5$ ?
(b) Find the acceleration of the particle at time $t=5$.
(c) Given that $x(t)$ is the position of the particle at time $t$ and that $x(0)=12$, find $x(3)$.
(d) Find the total distance traveled by the particle from $t=0$ to $t=6$.
(e) Find the average speed of the particle from $t=0$ to $t=6$.
(f) For what values of $t, 0 \leq t \leq 6$, is the particle’s instantaneous velocity the same as its average velocity on the closed interval $[0,6]$ ?
▶️Answer/Explanation
Solution
(a) $v(5)=(5)^2-3(5)-4>0$
Since $v(5)>0$, particle is moving to the right.
(b) $a(t)=v^{\prime}(t)=2 t-3$ So $a(5)=2(5)-3=7$.
(c) $\begin{aligned} & x(3)-x(0)=\int_0^3 v(t) d t=\int_0^3\left(t^2-3 t-4\right) d t \\ = & {\left[\frac{1}{3} t^3-\frac{3}{2} t^2-4 t\right]_0^3=-16.5 } \\ & x(3)=x(0)-16.5=12-16.5=-4.5\end{aligned}$
$\begin{aligned} & \text { (d) Total distance traveled }=\int_a^b|v(t)| d t=\int_0^6\left|t^2-3 t-4\right| d t \\ & =-\int_0^4\left(t^2-3 t-4\right) d t+\int_4^6\left(t^2-3 t-4\right) d t=\frac{56}{3}+\frac{38}{3}=\frac{94}{3}\end{aligned}$
(e) Average speed $=\frac{\text { distance traveled }}{\text { time }}=\frac{94 / 3}{6}=\frac{47}{9}$
$\begin{aligned} & \text { (f) Average velocity }=\frac{\text { displacement }}{\text { time }}=\frac{x(6)-x(0)}{6-0} \\ & =\frac{\int_0^6\left(t^2-3 t-4\right) d t}{6}=\frac{-6}{6}=-1 \\ & t^2-3 t-4=-1 \Rightarrow t^2-3 t-3=0 \quad \text { Since } t \geq 0, t=\frac{3+\sqrt{21}}{2} .\end{aligned}$
Example 2
A car is traveling on a straight road with velocity $80 \mathrm{ft} / \mathrm{sec}$ at time $t=0$. For $0 \leq t \leq 35$ seconds, the cars acceleration $a(t)$, in $\mathrm{ft} / \mathrm{sec}^2$, is the piecewise linear function defined by the graph above.
(a) Find $a(15)$ and $v(15)$.
(b) At what time, other than $t=0$, on the interval $0 \leq t \leq 35$, is the velocity of the car $80 \mathrm{ft} / \mathrm{sec}$ ?
(c) On the time interval $0 \leq t \leq 35$, what is the car’s absolute maximum velocity, in $\mathrm{ft} / \mathrm{sec}$, and at what time does it occur?
(d) On the time interval $0 \leq t \leq 35$, what is the car’s absolute minimum velocity, in $\mathrm{ft} / \mathrm{sec}$, and at what time does it occur?
▶️Answer/Explanation
Solution
(a) $a(15)=-20$
$
\begin{aligned}
& v(15)=v(0)+\int_0^{15} a(t) d t \\
& =v(0)+\int_0^{10} a(t) d t+\int_{10}^{15} a(t) d t \\
& =80+\frac{1}{2}(10+5) \cdot 20-\frac{1}{2}(5)(20) \\
& =180
\end{aligned}
$
$
v(0)=80
$
(b) At time $t=20$. Because the area above the $x$-axis from $t=0$ to $t=10$ is same as the area below the $x$-axis from $t=10$ to $t=20$.
(c) Check endpoints and the points where $a(t)$ changes signs.
$
\begin{aligned}
v(0) & =80 \\
v(10) & =v(0)+\int_0^{10} a(t) d t \quad \text { When } t=10, a(t) \text { changes signs. } \\
& =80+\frac{1}{2}(10+5) \cdot 20=230 \\
v(30) & =v(0)+\int_0^{30} a(t) d t \\
& =80+\frac{1}{2}(10+5) \cdot 20-\frac{1}{2}(20+5) \cdot 20 \quad \text { When } t=30, a(t) \text { changes signs. } \\
& =-20 \\
v(35) & =v(0)+\int_0^{35} a(t) d t \\
& =80+\frac{1}{2}(10+5) \cdot 20-\frac{1}{2}(20+5) \cdot 20+\frac{1}{2}(5)(10) \\
& =5
\end{aligned}
$
The car’s absolute maximum velocity is $230 \mathrm{ft} / \mathrm{sec}$ and it occurs at $t=10$.
(d) The car’s absolute minimum velocity is $-20 \mathrm{ft} / \mathrm{sec}$ and it occurs at $t=30$.
Example 3
- The acceleration of a particle moving along the $x$-axis at time $t$ is given by $a(t)=2 t-6$.
If at $t=1$, the velocity of the particle is 3 and its position is $\frac{1}{3}$, then the position $x(t)=$
(A) $\frac{t^3}{3}-6 t^2+5 t+\frac{1}{3}$
(B) $\frac{t^3}{3}-3 t^2+8 t-5$
(C) $\frac{t^3}{3}-6 t+9$
(D) $\frac{t^3}{3}-3 t^2+8 t-\frac{7}{3}$
▶️Answer/Explanation
Ans:B
Example 4
- The velocity of a particle moving along the $x$-axis at any time $t$ is given by $v(t)=3 e^{-t}-t$. What is the average speed of the particle over the time interval $0 \leq t \leq 3$ ?
(A) 0.873
(B) 1.096
(C) 1.273
(D) 1.482
▶️Answer/Explanation
Ans:D
Example 5
- A particle travels along a straight line with a velocity of $v(t)=e^t\left(t^2-5 t+6\right)$ meters per second. What is the average velocity of the particle over the time interval $0 \leq t \leq 5$ ?
(A) 58.602
(B) 64.206
(C) 79.351
(D) 86.448
▶️Answer/Explanation
Ans:D
Average Value of a Function
Definition of the Average Value of a Function
If $f$ is integrable on $[a, b]$, then the average value of $f$ on the interval is
$
\frac{1}{b-a} \int_a^b f(x) d x
$
The Mean Value Theorem for Definite Integrals
If $f$ is continuous on $[a, b]$, then there exists a number $c$ in $[a, b]$, such that
$
\int_a^b f(x) d x=f(c)(b-a)
$
In the figure when $f \geq 0$, the area of the rectangle, which is $f(c)(b-a)$, is equal to the area of the region bounded by the graph of $f$, the $x$-axis, and the vertical lines $x=a$ and $x=b$.
Example 1
- Find the average value of $f(x)=\frac{1}{2} x \cos \left(x^2\right)+x$ on the interval $[0, \sqrt{2 \pi}]$.
▶️Answer/Explanation
Solution
Average value
$
\begin{aligned}
& =\frac{1}{\sqrt{2 \pi}-0} \int_0^{\sqrt{2 \pi}}\left(\frac{1}{2} x \cos x^2+x\right) d x \\
& =\frac{1}{\sqrt{2 \pi}}\left[\int_0^{\sqrt{2 \pi}}\left(\frac{1}{2} x \cos x^2\right) d x+\int_0^{\sqrt{2 \pi}} x d x\right] \\
& =\frac{1}{2 \sqrt{2 \pi}} \int_0^{\sqrt{2 \pi}} x \cos x^2 d x+\frac{1}{\sqrt{2 \pi}} \int_0^{\sqrt{2 \pi}} x d x
\end{aligned}
$
$=\frac{1}{4 \sqrt{2 \pi}} \int_0^{2 \pi} \cos u d u+\frac{1}{2 \sqrt{2 \pi}} \int_0^{2 \pi} d u \quad$ Let $u=x^2$, then $d u=2 x d x . \Rightarrow \frac{1}{2} d u=x d x$.
$
=\frac{1}{4 \sqrt{2 \pi}}[\sin u]_0^{2 \pi}+\frac{1}{2 \sqrt{2 \pi}}[u]_0^{2 \pi}
$
If $x=0, u=0$ and if $x=\sqrt{2 \pi}, u=2 \pi$.
$\begin{aligned} & =\frac{1}{4 \sqrt{2 \pi}}[\sin 2 \pi-\sin 0]+\frac{1}{2 \sqrt{2 \pi}}[2 \pi-0] \\ & =\frac{\pi}{\sqrt{2 \pi}} \approx 1.253\end{aligned}$
Example 2
- Let $f$ be the function given by $f(x)=x^3-2 x+4$ on the interval $[-1,2]$. Find $c$ such that average value of $f$ on the interval is equal to $f(c)$.
▶️Answer/Explanation
Solution
Average value
$
\begin{aligned}
& =\frac{1}{b-a} \int_a^b f(x) d x=\frac{1}{2-(-1)} \int_{-1}^2\left(x^3-2 x+4\right) d x \\
& =\frac{1}{3}\left[\frac{x^4}{4}-x^2+4 x\right]_{-1}^2=4.25
\end{aligned}
$
Therefore, $f(c)=4.25$.
Use a graphing calculator to graph $y_1=x^3-2 x+4$ and $y_2=4.25$.
Find the points of intersection using a graphing calculator.
There are two points of intersection on the interval $[-1,2]$,
$(-.126,4.25)$ and $(1.473,4.25)$.
Therefore $c=-.126$ or $c=1.473$.
Example 3
- What is the average value of $f(x)=\sqrt{x}(4-x)$ on the closed interval $[0,4]$ ?
(A) $\frac{7}{3}$
(B) $\frac{21}{5}$
(C) $\frac{32}{15}$
(D) $\frac{35}{4}$
▶️Answer/Explanation
Ans:C
Example 4
- The graph of $y=f(x)$ consists of a semicircle and two line segments. What is the average value of $f$ on the interval $[0,8]$ ?
(A) $\frac{\pi+2}{4}$
(B) $\frac{\pi+3}{4}$
(C) $\pi+1$
(D) $\frac{\pi+6}{4}$
▶️Answer/Explanation
Ans:D
Example 5
The graph of $y=f(x)$ consists of three line segments as shown above. If the average value of $f$ on the interval $[0,5]$ is 1 what is the value of $k$ ?
(A) $\frac{3}{5}$
(B) $\frac{7}{10}$
(C) $\frac{4}{5}$
(D) $\frac{9}{10}$
▶️Answer/Explanation
Ans:C
Example 6
The function $f$ is continuous for $-4 \leq x \leq 4$. The graph of $f$ shown above consists of three line segments. What is the average value of $f$ on the interval $-4 \leq x \leq 4$ ?
(A) -1
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) 1
▶️Answer/Explanation
Ans:B
Length of a Curve (Distance Traveled Along a Curve)
If $f^{\prime}$ is continuous on the closed interval $[a, b]$, then the length of the curve $y=f(x)$ from $x=a$ to $x=b$ is
$L=\int_a^b \sqrt{1+\left[\frac{d y}{d x}\right]^2} d x=\int_a^b \sqrt{1+\left[f^{\prime}(x)\right]^2} d x$
If $g^{\prime}$ is continuous on the closed interval $[c, d]$, then the length of the curve $x=g(y)$ from $y=c$ to $y=d$ is
$
L=\int_c^d \sqrt{1+\left[\frac{d x}{d y}\right]^2} d y=\int_c^d \sqrt{1+\left[g^{\prime}(y)\right]^2} d y
$
Example 1
- Find the length of the curve $y=2 x^{3 / 2}+1$, from $x=1$ to $x=3$.
▶️Answer/Explanation
Solution
$\begin{aligned} & \frac{d y}{d x}=2 \cdot \frac{3}{2} x^{1 / 2}=3 x^{1 / 2} \\ &\left(\frac{d y}{d x}\right)^2=\left(3 x^{1 / 2}\right)^2=9 x \\ & L=\int_1^3 \sqrt{1+\left[\frac{d y}{d x}\right]^2} d x=\int_1^3 \sqrt{1+9 x} d x \\ &=\frac{2}{3} \cdot \frac{1}{9}\left[(1+9 x)^{3 / 2}\right]_1^3 \approx 8.633\end{aligned}$
Example2
- Let $R$ be the region bounded by the $y$-axis and the graphs of $y=x^2$ and $y=x+2$.
Find the perimeter of the region $R$.
▶️Answer/Explanation
Solution
The $y$-intercept of the line is $A(0,2)$. We can find the point of intersection of the two graphs by solving $y=x^2$ and $y=x+2$ simultaneously for $x$. The point of intersection is $B(2,4)$.
The perimeter of the region $R$
$=O A+A B+$ length of the curve from $O$ to $B$
$=2+\sqrt{(4-2)^2+(2-0)^2}+\int_0^2 \sqrt{1+(2 x)^2} d x \quad \frac{d y}{d x}=2 x$
$
\begin{aligned}
& =2+2 \sqrt{2}+4.647 \\
& =9.475
\end{aligned}
$
Use a graphing calculator to find
$
\text { the value of } \int_0^2 \sqrt{1+(2 x)^2} d x \text {. }
$
Example 3
- What is the length of the curve of $y=\frac{1}{3}\left(x^2+2\right)^{3 / 2}$ from $x=1$ to $x=2$ ?
(A) $\frac{8}{3}$
(B) $\frac{10}{3}$
(C) 4
(D) $\frac{14}{3}$
▶️Answer/Explanation
Ans:B
Example 4
- Which of the following integrals gives the length of the graph of $y=\ln (\sin x)$ between $x=\frac{\pi}{3}$ to $x=\frac{2 \pi}{3}$ ?
(A) $\int_{\pi / 3}^{2 \pi / 3} \csc ^2 x d x$
(B) $\int_{\pi / 3}^{2 \pi / 3} \sqrt{1+\cot x} d x$
(C) $\int_{\pi / 3}^{2 \pi / 3} \csc x d x$
(D) $\int_{\pi / 3}^{2 \pi / 3} \sqrt{1+\csc ^2 x} d x$
▶️Answer/Explanation
Ans:C
Example 5
- Which of the following integrals gives the length of the graph of $y=\frac{1}{3} x^{3 / 2}-x^{1 / 2}$ between $x=1$ to $x=4 ?$
(A) $\frac{1}{2} \int_1^4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
(B) $\frac{1}{2} \int_1^4\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right) d x$
(C) $\frac{1}{2} \int_1^4\left(1+\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
(D) $\frac{1}{2} \int_1^4\left(1+\sqrt{x}-\frac{1}{\sqrt{x}}\right) d x$
▶️Answer/Explanation
Ans:A
Example6
What is the length of the curve of $y=\ln \left(x^2+1\right)-x$ from $x=0$ to $x=3$ ?
(A) 1.026
(B) 1.826
(C) 2.227
(D) 3.135
▶️Answer/Explanation
Ans:D
Example6
- If the length of a curve from $(0,-3)$ to $(3,3)$ is given by $\int_0^3 \sqrt{1+\left(x^2-1\right)^2} d x$, which of the following could be an equation for this curve?
(A) $y=\frac{x^3}{3}-\frac{x}{3}-3$
(B) $y=\frac{x^3}{3}-3 x-3$
(C) $y=\frac{x^3}{3}-x-3$
(D) $y=\frac{x^3}{3}+x-3$
▶️Answer/Explanation
Ans:C
Example7
- If $F(x)=\int_1^{x^2} \sqrt{t+1} d t$, what is the length of the curve of from $x=1$ to $x=2$ ?
(A) $\frac{8}{3}$
(B) $\frac{10}{3}$
(C) $\frac{15}{3}$
(D) $\frac{17}{3}$
▶️Answer/Explanation
Ans:D