적분법 전체 문제

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 $0-(y^2+y)$.

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

We need to figure out which curve is on top, when. To do this, set $h(x) = 3x - x\sqrt{25-x^2}$. If $h(x) > 0$, then $y=3x$ is the top curve; if $h(x)<0$, then $y=x\sqrt{25-x^2}$ is the top curve.

$$\begin{aligned} h(x) &= 3x - x\sqrt{25-x^2} = x\left[3-\sqrt{25-x^2}\right] \\ \text{We only care about values of $x$ in $[0,4]$, so $x$ is nonnegative. Then $h(x)$ is positive when:} \\ 3&> \sqrt{25-x^2}\\ 9&> 25-x^2\\ x^2 & > 16\\ x&> 4 \end{aligned}$$

That is, $h(x)$ is never positive over the interval $[0,4]$. So, $y = x \sqrt{25-x^2}$ lies above $y=3x$ for all $0\le x\le 4$.

The area we need to calculate is therefore:

$$\begin{aligned} A &= \int_0^4 \left[x \sqrt{25-x^2} - 3x\right]\,\mathrm{d}x \\ &= \int_0^4 x \sqrt{25-x^2}\,\mathrm{d}x - \int_0^4 3x\,\mathrm{d}x \\&= A_1 - A_2. \end{aligned}$$

To evaluate $A_1$, we use the substitution $u(x) = 25-x^2$, for which $\mathrm{d}u = u'(x)\,\mathrm{d}x= -2x\,\mathrm{d}x$; and $u(4)=25-4^2=9$ when $x=4$, while $u(0)=25-0^2=25$ when $x=0$. Therefore

$$\begin{aligned} A_1 &= \int_{x=0}^{x=4} x \sqrt{25-x^2}\,\mathrm{d}x = -\frac{1}{2} \int_{u=25}^{u=9} \sqrt{u}\,\mathrm{d}u = \left[-\frac{1}{3} u^{3/2} \right]_{25}^{9} = \frac{125 - 27}{3} = \frac{98}{3} \end{aligned}$$

For $A_2$ we use the antiderivative directly:

$$A_2 = \int_0^4 3x\,\mathrm{d}x =\left[ \frac{3x^2}{2} \right]_0^4 = 24$$

Therefore the total area is:

$$\begin{aligned} A = \frac{98}{3} - 24 = \frac{26}{3} \end{aligned}$$

Remember our rule: $\int u \mathrm{d}v = uv - \int v \mathrm{d}u$. So, we take $u$ and use it to make $\mathrm{d}u$–that is, we differentiate it. We take $\mathrm{d}v$ and use it to make $v$–that is, we antidifferentiate it.

This integrand is the product of two functions, with no obvious substitution. So, let's try integration by parts, with one part $e^x$ and one part $x^3$.

$$\begin{array}{c | l | l |} &&\\ \text{Option 1:\quad}& u=e^x &\mathrm{d}u=e^x ~\mathrm{d}x\\[10pt] ~& \mathrm{d}v=x^3~\mathrm{d}x&v=\frac{1}{4}x^4\\[10pt] &&\\ \text{Option 2:\quad}& u=x^3&\mathrm{d}u=3x^2~\mathrm{d}x\\[10pt] ~& \mathrm{d}v=e^x~ \mathrm{d}x&v=e^x\\[10pt] \end{array}$$

At first glance, multiplying the red functions and multiplying the blue functions give largely equivalent integrands to what we started with–none of them with obvious antiderivatives. In previous questions, we were able to choose $u=x$, and then $\mathrm{d}u=\mathrm{d}x$, so the “$x$" in the integrand effectively went away. Here, we see that choosing $u=x^3$ will lead to $\mathrm{d}u=3x^2\mathrm{d}x$, which has a lower power. If we repeatedly perform integration by parts, choosing $u$ to be the power of $x$ each time, then after a few iterations it should go away, because the third derivative of $x^3$ is a constant.

So, we start with Option 2: $u=x^3$, $\mathrm{d}v=e^x\mathrm{d}x$, $\mathrm{d}u=3x^2\mathrm{d}x$, and $v = e^x$.

$$\begin{aligned} \int x^3 e^x \mathrm{d}x&= \underbrace{x^3}_{u}\underbrace{e^x}_{v} - \int \underbrace{e^x}_v \cdot\underbrace{ 3x^2 \mathrm{d}x}_{\mathrm{d}u}\\ &= {x^3}{e^x} -3 \int {e^x} \cdot x^2 \mathrm{d}x \\ \text{Now, we take $u=x^2$ and $\mathrm{d}v=e^x\mathrm{d}x$, so $\mathrm{d}u=2x\mathrm{d}x$ and $v=e^x$. We're only using integration by parts on the actual integral--the rest of the function stays the way it is.} \\ &=x^3e^x -3 \left[\underbrace{x^2e^x}_{uv} - \int \underbrace{e^x}_{v} \cdot \underbrace{2x\mathrm{d}x}_{\mathrm{d}u}\right] \\ &=x^3e^x - 3x^2e^x + 6\int x e^x\mathrm{d}x \\ \text{Continuing, we take $u=x$ and $\mathrm{d}v=e^x\mathrm{d}x$, so $\mathrm{d}u=\mathrm{d}x$ and $v=e^x$. This is the step where the polynomial part of the integrand finally disappears.} \\ &=x^3e^x - 3x^2e^x + 6\left[\underbrace{xe^x}_{uv} - \int \underbrace{ e^x}_v \underbrace{\mathrm{d}x}_{\mathrm{d}u}\right]\\ &=x^3e^x-3x^2e^x+6xe^x - 6e^x+C \\ &=e^x\left(x^3-3x^2+6x - 6\right)+C \end{aligned}$$

Let's check that this makes sense: the derivative of $e^x\left(x^3-3x^2+6x - 6\right)+C$ should be $x^3e^x$. We differentiate using the product rule.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^x\left(x^3-3x^2+6x - 6\right)+C\right\} &=e^x\left(x^3-3x^2+6x - 6\right)+e^x\left(3x^2-6x+6\right)\\ &=e^x\left(x^3-3x^2+3x^2+6x-6x-6+6\right)=x^3e^x \end{aligned}$$

Remark: In order to be technically correct in our antidifferentiation, we should add the $+C$ as soon as we do the first integration by parts. However, when we are using integration by parts, we usually end up evaluating an integral at the end, and we add the $+C$ at that point. Since the $+C$ comes up eventually, it is common practice to not clutter our calculations with it until the end.

The integrand is the product of two functions, without an obvious substitution, so let's see what integration by parts can do for us.

$$\begin{array}{c | l | l |} &&\\ \text{Option 1:\quad}& u=x^2 &\mathrm{d}u=2x ~\mathrm{d}x\\[10pt] ~& \mathrm{d}v=\sin x~\mathrm{d}x&v=-\cos x\\[10pt] &&\\ \text{Option 2:\quad}& u=\sin x&\mathrm{d}u=\cos x~\mathrm{d}x\\[10pt] ~& \mathrm{d}v=x^2~ \mathrm{d}x&v=\frac{1}{3}x^3\\[10pt] \end{array}$$

Neither option gives us something immediately integrable, but Option 1 replaces our $x^2$ term with a lower power of $x$. If we repeatedly apply integration by parts, we can reduce this power to zero. So, we start by choosing $u=x^2$ and $\mathrm{d}v=\sin x\mathrm{d}x$, so $\mathrm{d}u=2x\mathrm{d}x$ and $v=-\cos x$.

$$\begin{aligned} \int x^2\sin x~\mathrm{d}x &=\underbrace{ -x^2\cos x}_{uv} +\underbrace{ \int 2x\cos x \mathrm{d}x}_{-v\mathrm{d}u}\\ &=-x^2\cos x + 2 \int x \cos x \mathrm{d}x \\ \text{Using integration by parts again, we want to be differentiating (not antidifferentiating) $x$, so we choose $u=x$, $\mathrm{d}v=\cos x \mathrm{d}x$, and then $\mathrm{d}u=\mathrm{d}x$ ($x$ went away!), $v=\sin x$.} \\ &=-x^2\cos x + 2\left[\underbrace{x\sin x}_{uv}-\int \underbrace{\sin x \mathrm{d}x}_{v\mathrm{d}u} \right] \\&=-x^2\cos x + 2x\sin x+2\cos x +C\\ &=(2-x^2)\cos x + 2x\sin x +C \end{aligned}$$

This is your friendly reminder that to a person with a hammer, everything looks like a nail. The integral in the problem is a classic example of an integral to solve using substitution. We have an “inside function," $x^2+1$, whose derivative shows up multiplied to the rest of the integrand. We take $u=x^2+1$, then $\mathrm{d}u=2x\mathrm{d}x$, so $$\int 2xe^{x^2+1}\mathrm{d}x = \int e^u\mathrm{d}u=e^u+C = e^{x^2+1}+C$$

We integrate by parts, using $u = \arctan(2y)$, $\mathrm{d}v = 4y \,\mathrm{d}y$, so that $v=2y^2$ and $\mathrm{d}u= \frac{2\, \mathrm{d}y}{1+(2y)^2}$:

$$\begin{aligned} \int 4y\arctan(2y) \,\mathrm{d}y = \underbrace{2y^2\arctan(2y)}_{uv} - \int\underbrace{ \frac{4y^2}{(2y)^2+1} \,\mathrm{d}y}_{v\mathrm{d}u} \end{aligned}$$

The integrand $\frac{4y^2}{(2y)^2+1}$ is a rational function. So the remaining integral can be evaluated using the method of partial fractions, starting with long division. But it is easier to just notice that $\frac{4y^2}{4y^2+1} = \frac{4y^2+1}{4y^2+1} - \frac{1}{4y^2+1}$. We therefore have:

$$\begin{aligned} \int \frac{4y^2}{4y^2+1} \,\mathrm{d}y = \int \left(1 - \frac{1}{4y^2+1}\right)\,\mathrm{d}y = y - \frac12\arctan(2y) + C \end{aligned}$$

The final answer is then

$$\begin{aligned} \int 4y\arctan(2y) \,\mathrm{d}y = 2y^2\arctan(2y) - y + \frac{1}{2}\arctan(2y) + C \end{aligned}$$

It's not obvious where to start, but in general it's nice to have the arguments of our trig functions the same. So, we use the identity $\sin(2x)=2\sin x \cos x$.

$$\begin{aligned} \int e^{\cos x}\sin(2x)\mathrm{d}x&=2\int e^{\cos x}\cos x \sin x\,\mathrm{d}x \\ \text{Now we can use the substitution $w=\cos x$, $\mathrm{d}w=-\sin x \mathrm{d}x$.} \\ &=-2\int we^w\mathrm{d}w \\ \text{From here the integral should look more familiar. We can use integration by parts with $u=w$, $\mathrm{d}v = e^w\mathrm{d}w$, $\mathrm{d}u=\mathrm{d}w$, and $v=e^w$.} \\ &=-2\left[\underbrace{we^w}_{uv} - \int \underbrace{e^w\mathrm{d}w}_{v\mathrm{d}u}\right]\\ &=2e^w\left[1-w\right]+C\\ &=2e^{\cos x}[1-\cos x]+C \end{aligned}$$

We've got an integrand that consists of several functions multiplied together, and no obvious substitution. So, we think about integration by parts. We know an antiderivative for $\frac{1}{(1-x)^2}$, because we know $\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{1-x}=\frac{1}{(1-x)^2}$. So let's try $\mathrm{d}v = \frac{\mathrm{d}x}{(1-x)^2}$ and $u=xe^{-x}$. Then $v=\frac{1}{1-x}$ and $\mathrm{d}u = (1-x)e^{-x}\,\mathrm{d}x$. So, by integration by parts,

$$\begin{aligned} \int\underbrace{xe^{-x}}_{u}\underbrace{ \frac{\mathrm{d}x}{(1-x)^2}}_{\mathrm{d}v} &= \underbrace{\frac{xe^{-x}}{1-x}}_{uv} -\int\underbrace{\frac{1}{1-x}}_{v}\underbrace{(1-x)e^{-x}\,\mathrm{d}x}_{\mathrm{d}u} \\ &=\frac{xe^{-x}}{1-x} -\int e^{-x}\,\mathrm{d}x \\ &=\frac{xe^{-x}}{1-x} + e^{-x} + C =\frac{e^{-x}}{1-x} + C \end{aligned}$$