적분법 전체 문제

Setting $u=x^3+1$, we have $\mathrm{d}u = 3x^2\,\mathrm{d}x$ and so

$$\begin{aligned} \int \frac{ x^2\,\mathrm{d}x}{{( x^3+1)}^{101}} &= \int \frac{ \mathrm{d}u/3}{ u^{101}}\\ &=\frac{1}{3}\int u^{-101}\ \mathrm{d}u\\ &=\frac{1}{3}\cdot\dfrac{u^{-100}}{-100}\\ &= -\frac{1}{3\times 100 u^{100}} + C\\ &=-\frac{1}{300{(x^3+1)}^{100}} + C \end{aligned}$$

Setting $u=\log x$, we have $\mathrm{d}u = \frac{1}{x}\,\mathrm{d}x$ and so

$$\int_{e}^{e^4} \frac{ \mathrm{d}x}{{blue}x\cdot \log x} = \int_{x=e}^{x=e^4} \frac1{ \log x} \cdot \frac{1}{x}\,\mathrm{d}x = \int_{u=1}^{u=4} \frac{1}{{red}u}\, \mathrm{d}u,$$

since $u=\log(e)=1$ when $x=e$ and $u=\log(e^4)=4$ when $x=e^4$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{1}^{4} \frac{1}{u}\, \mathrm{d}u = \Big[\log |u| \Big]_{1}^{4} = \log 4 - \log 1 = \log 4.$$

Setting $u=1+\sin x$, we have $\mathrm{d}u = \cos x~\mathrm{d}x$ and so

$$\int_{0}^{\pi/2} \frac{ \cos x} { 1+\sin x} \mathrm{d}x = \int_{x=0}^{x=\pi/2} \frac{1}{ 1+\sin x}\, \cos x ~\mathrm{d}x = \int_{u=1}^{u=2} \frac{ \mathrm{d}u}{{red}u}$$

since $u=1+\sin 0=1$ when $x=0$ and $u=1+\sin(\pi/2)=2$ when $x=\pi/2$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{u=1}^{u=2} \frac{\mathrm{d}u}{u} = \Big[\log|u| \Big]_{1}^{2} = \log 2$$

Setting $u=\sin x$, we have $\mathrm{d}u = \cos x~\mathrm{d}x$ and so

$$\int_{0}^{\pi/2} \cos x \cdot (1+ \sin^2 {red}x) \mathrm{d}x = \int_{x=0}^{x=\pi/2} (1+ \sin^2 x)\cdot \cos x ~\mathrm{d}x = \int_{u=0}^{u=1} (1+ u^2) \, \mathrm{d}u,$$

since $u=\sin 0=0$ when $x=0$ and $u=\sin(\pi/2)=1$ when $x=\pi/2$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{0}^{1} (1+u^2) \,\mathrm{d}u = \left[u+\frac{u^3}{3} \right]_{0}^{1} =\left(1+\frac{1}{3}\right) -0 = \frac{4}{3}.$$

Substituting $t=x^2-x$, $\mathrm{d}t = (2x-1)\,\mathrm{d}x$ and noting that $t=0$ when $x=1$ and $t=6$ when $x=3$,

$$\begin{aligned} \int_1^3 (2x-1)e^{ x^2-x} \mathrm{d}x &= \int_0^6 e^{{red}t}\ \mathrm{d}t =\big[e^t\big]_0^6 =e^6-1 \end{aligned}$$

We use the substitution $u=4-x^2$, for which $\mathrm{d}u=-2x\,\mathrm{d}x$ :

$$\begin{aligned} \int \frac{x^2-4}{\sqrt{4-x^2}}\,x\,\mathrm{d}x &=\int \frac{1}{2}\cdot\frac{ 4-x^2}{\sqrt{ 4-x^2}} ({-}2x)\,\mathrm{d}x \\ &=\frac{1}{2} \int \frac{{red}u}{\sqrt{{red}u}}\, \mathrm{d}u \\ &=\frac{1}{2}\int \sqrt{u}\,\mathrm{d}u \\ &=\frac{1}{2}\frac{u^{3/2}}{3/2}+C\\ &=\frac{1}{3}(4-x^2)^{3/2}+C \end{aligned}$$

• Solution 1: If we let $u=\sqrt{\log x}$, then $\mathrm{d}u=\dfrac{1}{2x\sqrt{\log x}}\ \mathrm{d}x$, and:

$$\begin{aligned} \int \dfrac{e^{ \sqrt{\log x}}}{ 2x\sqrt{\log x}}\ \mathrm{d}x&=\int e^{{red}u}\ \mathrm{d}u=e^u+C=e^{\sqrt{\log x}}+C \end{aligned}$$

• Solution 2: In Solution 1, we made a pretty slick choice. We might have tried to work with something a little less convenient. For example, it's not unnatural to think that $u=\log x$, $\mathrm{d}u=\dfrac{1}{x}\ \mathrm{d}x$ would be a good choice. In that case:

$$\begin{aligned} \int \dfrac{e^{\sqrt{ \log x}}}{2 x\sqrt{ \log x}}\ \mathrm{d}x&= \int \frac{e^{\sqrt{{red}u}}}{2\sqrt{{red}u}} \mathrm{d}u \\ \text{Now, we should be able to see that $ w=\sqrt{u}$, $ \mathrm{d}w = \dfrac{1}{2\sqrt{u}}\ \mathrm{d}u$ is a good choice:} \\ \int\frac{e^{ \sqrt{u}}}{ 2\sqrt{u}}\ \mathrm{d}u &=\int e^{{orange}w}\ \mathrm{d}w\\ &=e^{\sqrt{u}}+C\\ &=e^{\sqrt{\log x}}+C \end{aligned}$$

The area between the curve $y= \frac{1}{(2x-4)^2}$ and the $x$-axis, with $x$ running from $a=0$ to $b=1$, is exactly the definite integral of $\frac{1}{(2x-4)^2}$ with limits $0$ and $1$.

$$\begin{aligned} \text{Area}&=\int_0^1 \frac{\mathrm{d}x}{(2x-4)^2}&u=2x-4,\quad\mathrm{d}u=2\ \mathrm{d}x\\ &=\frac{1}{2}\int_{-4}^{-2}\frac{1}{u^2}\mathrm{d}u = \frac{1}{2}\left[\frac{-1}{u}\right]_{u=-4}^{u=-2}\\ &=\frac{1}{2}\Big[\frac{1}{2}-\frac{1}{4}\Big] =\frac{1}{8} \end{aligned}$$

If the curves $y=f(x)=x$ and $y=g(x)=3x-x^2$ intersect at $(x,y)$, then

$$\begin{aligned} 3x-x^2&=y=x\\ x^2-2x&=0\\ x(x-2)&=0\\ x=0 \quad &\text{or} \quad x=2 \end{aligned}$$

Furthermore, $g(x)-f(x) = 2x-x^2 = x(2-x)$ is positive for all $0\le x\le 2$. That is, the curve $y=3x-x^2$ lies above the line $y=x$ for all $0\le x\le 2$.

We therefore evaluate the integral:

$$\begin{aligned} \int_0^2 \big[ (3x-x^2) - x \big] \,\mathrm{d}x = \int_0^2 [2x-x^2]\,\mathrm{d}x = \bigg[x^2 - \frac{x^3}{3}\bigg]^{2}_{0} = \bigg[ 4-\frac{8}{3} \bigg] -0 = \frac{4}{3} \end{aligned}$$

For our computation, we will need an antiderivative of $x^2\sqrt{x^3+1}$, which can be found using the substitution $u=x^3+1$, $\mathrm{d}u = 3x^2\,\mathrm{d}x$:

$$\begin{aligned} \int x^2\sqrt{x^3+1} \, \mathrm{d}x = \int \sqrt u \cdot \frac13\,\mathrm{d}u = \frac13\int u^{1/2}\,\mathrm{d}u = \frac13\cdot \frac{u^{3/2}}{3/2}+C = \frac29(x^3+1)^{3/2} + C. \end{aligned}$$

The two functions $f(x)$ and $g(x)$ are clearly equal at $x=0$. If $x\ne0$, then the functions are equal when

$$\begin{aligned} 3x^2 &= x^2\sqrt{x^3+1} \\ 3 &= \sqrt{x^3+1} \\ 9 &= x^3+1 \\ 8 &= x^3 \\ 2 &= x. \end{aligned}$$

The function $g(x)=3x^2$ is the larger of the two on the interval $[0,2]$, as can be seen by plugging in $x=1$, say, or by observing that when $x$ is very small $f(x)=x^2\sqrt{x^3+1}\approx x^2$ and $g(x)=3x^2$.

The area in question is therefore:

$$\begin{aligned} \int_0^2 \big( 3x^2 - x^2\sqrt{x^3+1} \big) \, \mathrm{d}x &= \bigg( {x^3} - \frac29(x^3+1)^{3/2} \bigg) \bigg|_0^2 \\ &= \bigg( 2^3 - \frac2 9(2^3+1)^{3/2} \bigg) - \bigg( 0^3 - \frac29(0^3+1)^{3/2} \bigg) \\ &= \bigg( 8 - 6 \bigg) - \bigg( 0 - \frac 2 9 \bigg) =\frac{20}9. \end{aligned}$$

First, let's figure out what our curve $x=y^2+y=y(y+1)$ looks like.

• The curve intercepts the $y$-axis when $y=0$ and $y=-1$.
• The $x$-values of the curve are negative when $-1<y<0$, and positive elsewhere.

This leads to the figure below. We're evaluating the area from $y=-1$ to $y=0$. Since $y^2+y$ is negative there, the length of our (horizontal) slices are $0-(y^2+y)$.

$$\begin{aligned} \text{Area}=\int_{-1}^0\big(0-(y^2+y)\big)\,\mathrm{d}y = -\bigg[\frac{y^3}{3}+\frac{y^2}{2}\bigg]_{-1}^0 =-\frac13+\frac12 =\frac{1}{6}\quad \smash{  } \end{aligned}$$

Let's begin by sketching our region. Note that $y=\sqrt{1-x^2}$ and $y=\sqrt{9-x^2}$ are the top halves of circles centred at the origin with radii 1 and 3, respectively.

Our region is the difference of two quarter-circles, so we find its area using geometry: $$\text{Area}=\frac{1}{4}\left(\pi\cdot 3^2\right)-\frac{1}{4}\left(\pi\cdot 1^2\right)=2\pi$$

The curves $y=1/x$ and $3x+3y=10$, i.e. $y =\frac{10}{3}-x$ intersect when

$$\begin{aligned} \frac{1}{x} = \frac{10}{3}-x &\iff 3 = 10x-3x^2 \iff 3x^2-10x+3=0 \\ &\iff(3x-1)(x-3)=0 \\ &\iff x=3\,,\,\frac{1}{3} \end{aligned}$$

enterline

When the region is rotated about the $x$–axis, the vertical strip in the figure above sweeps out a washer with thickness $\mathrm{d}x$, outer radius $T(x)=\frac{10}{3}-x$ and inner radius $B(x)=\frac{1}{x}$. This washer has volume

$$\pi\big(T(x)^2- B(x)^2\big)\,\mathrm{d}x = \pi\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\mathrm{d}x$$

Hence the volume of the solid is

$$\begin{aligned} \pi\int_{1/3}^3\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\mathrm{d}x &=\pi\Big[\frac{100x}{9}-\frac{10}{3}x^2+\frac{1}{3}x^3 +\frac{1}{x}\Big]_{1/3}^3 \\ &=\pi\Big[\frac{38}{3}-\frac{514}{3^4}\Big] = \pi\frac{512}{81} \end{aligned}$$