Integrals and the Fundamental Theorem: practice solutions, College

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Practice solutions College : Integrals and the Fundamental Theorem — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

Sommaire

1 Checking an antiderivative ★★★

Differentiate $F$: $F'(x)=\dfrac{4x^3}{4}-3=x^3-3=f(x)$.

Since $F'=f$, $F$ is an antiderivative of $f$.

2 A general antiderivative ★★★

Use the power rule term by term: $\int 8x^3dx=2x^4$, $\int 6x^2dx=2x^3$, $\int 2x\,dx=x^2$, $\int 7\,dx=7x$.

So $\displaystyle\int f(x)\,dx=2x^4-2x^3+x^2-7x+C$.

3 A first definite integral ★★★

An antiderivative is $F(x)=x^2+3x$.

$F(4)=16+12=28$ and $F(1)=1+3=4$, so the integral equals $28-4=24$.

4 Using the properties ★★★

By linearity: $2(7)+5(-2)=14-10=4$.
Swapping the limits changes the sign: $-7$.
$7-(-2)=9$.

5 Left and right sums ★★★

Here $\Delta x=1$. Left endpoints are 0, 1, 2: $L_3=(0+2+4)\times1=6$. Right endpoints are 1, 2, 3: $R_3=(2+4+6)\times 1=12$.

The region is a triangle with base 3 and height 6, so the exact area is $\tfrac12\cdot3\cdot6=9$, which lies between 6 and 12 (it is their average).

6 Fixing the constant ★★★

Integrate: $F(x)=3x^2+x+C$. Then $F(2)=12+2+C=14+C$.

From $14+C=10$ we get $C=-4$, so $F(x)=3x^2+x-4$.

7 True or false? ★★★

True. $F(x)=\dfrac{x^4}{4}$ gives $F(2)-F(-2)=4-4=0$. (The graph of $x^3$ is symmetric about the origin, so the signed areas cancel.)
False. The left side is $\int_0^2 1\,dx=2$, while the right side is $2\times 2=4$. An integral of a product is not the product of integrals.

8 Sine and the natural log ★★★

$\big[-\cos x\big]_0^{\pi}=-\cos\pi+\cos0=1+1=2$.
$\big[\ln|x|\big]_1^{e}=\ln e-\ln 1=1-0=1$.

9 A midpoint estimate ★★★

$\Delta x=1$; the midpoints are 0.5, 1.5, 2.5, 3.5. The heights are $1.25,\ 3.25,\ 7.25,\ 13.25$, which sum to 25.

So $M_4=25\times1=25$. Exactly: $\left[\dfrac{x^3}{3}+x\right]_0^4=\dfrac{64}{3}+4=\dfrac{76}{3}\approx25.33$. The error is $\dfrac13\approx0.33$.

10 Area of a tunnel opening ★★★

The parabola meets the axis at $x=\pm3$. Area $=\displaystyle\int_{-3}^{3}(9-x^2)\,dx=\left[9x-\dfrac{x^3}{3}\right]_{-3}^{3}=(27-9)-(-27+9)=36$.

The cross-section is 36 m$^2$, about $36\times10.76\approx387$ ft$^2$.

11 Filling a water tank ★★★

The net change is $\displaystyle\int_0^5(12-2t)\,dt=\big[12t-t^2\big]_0^5=60-25=35$ gallons.

The tank then holds $80+35=115$ gallons, about $115\times3.785\approx435$ liters.

12 Two quick substitutions ★★★

Let $u=x^2$, $du=2x\,dx$. Then $\int\cos u\,du=\sin u+C=\sin(x^2)+C$.
Let $u=3x+2$, $du=3\,dx$, so $dx=\tfrac13du$. Then $\tfrac13\int u^4du=\dfrac{u^5}{15}+C=\dfrac{(3x+2)^5}{15}+C$.

13 Substitution with new limits ★★★

Let $u=x^2+1$, so $du=2x\,dx$ and $x\,dx=\tfrac12du$. The limits become $u=1$ and $u=2$.

$\displaystyle\tfrac12\int_1^2\dfrac{du}{u}=\tfrac12\big[\ln u\big]_1^2=\tfrac12\ln2\approx0.347$.

14 Differentiating an integral ★★★

By Part 1 of the Fundamental Theorem, $g'(x)=x^3-5$. Hence $g'(3)=27-5=22$.

An integral from 2 to 2 is zero, so $g(2)=0$.

15 The chain rule meets the FTC ★★★

By the chain version of Part 1, $h'(x)=\sqrt{1+(x^2)^2}\cdot 2x=2x\sqrt{1+x^4}$. So $h'(1)=2\sqrt2\approx2.83$.
Swap the limits: $k(x)=-\int_5^x\sin(t^2)\,dt$, so $k'(x)=-\sin(x^2)$.

16 A drone changes direction ★★★

Factor: $v(t)=(t-2)(t-4)$, positive on $[0,2]$, negative on $[2,4]$, positive on $[4,5]$. A position function is $s(t)=\dfrac{t^3}{3}-3t^2+8t$.

$s(0)=0$, $s(2)=\dfrac{20}{3}$, $s(4)=\dfrac{16}{3}$, $s(5)=\dfrac{20}{3}$.

Displacement: $s(5)-s(0)=\dfrac{20}{3}\approx6.67$ m. Distance: $\dfrac{20}{3}+\left|\dfrac{16}{3}-\dfrac{20}{3}\right|+\left(\dfrac{20}{3}-\dfrac{16}{3}\right)=\dfrac{28}{3}\approx9.33$ m.

17 Sine cubed times cosine ★★★

Let $u=\sin x$, $du=\cos x\,dx$. When $x=0$, $u=0$; when $x=\pi/2$, $u=1$.

$\displaystyle\int_0^1u^3\,du=\left[\dfrac{u^4}{4}\right]_0^1=\dfrac14$.

18 Logarithm substitution ★★★

Let $u=\ln x$, $du=\dfrac{dx}{x}$. The limits become $\ln1=0$ and $\ln e^2=2$.

$\displaystyle\int_0^2u\,du=\left[\dfrac{u^2}{2}\right]_0^2=2$.

19 Absolute value integral ★★★

$|x-1|=1-x$ on $[-1,1]$ and $x-1$ on $[1,3]$. By additivity,

$\displaystyle\int_{-1}^{1}(1-x)\,dx=\left[x-\dfrac{x^2}{2}\right]_{-1}^{1}=\dfrac12-\left(-\dfrac32\right)=2$ and $\displaystyle\int_1^3(x-1)\,dx=\left[\dfrac{x^2}{2}-x\right]_1^3=\dfrac32-\left(-\dfrac12\right)=2$.

The total is $4$. (Each piece is a triangle with base 2 and height 2.)

20 Marginal cost ★★★

The increase is $C(200)-C(100)=\displaystyle\int_{100}^{200}(0.06q+4)\,dq=\big[0.03q^2+4q\big]_{100}^{200}$.

$=(1{,}200+800)-(300+400)=2{,}000-700=1{,}300$. The cost rises by \$1,300.

21 From Riemann sums to an integral ★★★

With $\Delta x=\tfrac1n$ and $x_i=\tfrac in$: $R_n=\displaystyle\sum_{i=1}^n\dfrac{i^2}{n^2}\cdot\dfrac1n=\dfrac{1}{n^3}\cdot\dfrac{n(n+1)(2n+1)}{6}=\dfrac{(n+1)(2n+1)}{6n^2}$.

For $n=10$: $R_{10}=\dfrac{11\cdot21}{600}=0.385$.

As $n\to\infty$, $\dfrac{(n+1)(2n+1)}{6n^2}\to\dfrac{2}{6}=\dfrac13$, which equals $\displaystyle\int_0^1x^2\,dx$, as the Fundamental Theorem predicts.