적분법 전체 문제

$-\cos x+C$

$F(x)=-\cos x,F\left(\frac{\pi }{2}\right)-F(0)=1$

$\int \frac{dx}{2x}=\frac{1}{2}\ln |x|+C$

$\sqrt{{x}^{2}+1}+C$

$2\sqrt{26}$

$-\frac{{e}^{-{x}^{2}}}{2}+C$

The reasoning is not sound: when we do a substitution, we need to take care of the differential ($\mathrm{d}x$). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here's the way to do it:

Problem: Evaluate $\displaystyle\int (2x+1)^2 \mathrm{d}x$.

Work: We use the substitution $u=2x+1$. Then $\mathrm{d}u=2\mathrm{d}x$, so $\mathrm{d}x = \frac{1}{2}\mathrm{d}u$:

$$\begin{aligned} \int (2x+1)^2 \mathrm{d}x&=\int u^2\cdot \frac{1}{2}\ \mathrm{d}u\\ &=\frac{1}{6}u^3+C\\ &=\frac{1}{6}\left(2x+1\right)^3+C \end{aligned}$$

Perhaps shorter ways exist, but the reasoning here is valid.

Problem: Evaluate $\displaystyle\int_{0}^{\pi/4} x\tan (x^2) \ \mathrm{d}x$.

Work: We begin with the substitution $u=x^2$, $\mathrm{d}u = 2x\mathrm{d}x$:

If $u=x^2$, then $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x$, so indeed $\mathrm{d}u=2x\mathrm{d}x$.

$$\begin{aligned} \int_{0}^{\pi/4} x\tan (x^2) \ \mathrm{d}x&= \int_{0}^{\pi/4} \frac{1}{2}\tan(x^2)\cdot 2x\mathrm{d}x&\text{algebra}\\ &=\int_{0}^{\pi^2/16} \frac{1}{2}\tan u\ \mathrm{d}u \\ \text{\text{Every piece is changed from $x$ to $u$: integrand, differential, limits.}} \\ &=\frac{1}{2}\int_{0}^{\pi^2/16} \dfrac{\sin u}{\cos u}\mathrm{d}u & \tan u = \frac{\sin u}{\cos u}\\ \\ \text{Now we use the substitution $v=\cos u$, $\mathrm{d}v=-\sin u \ \mathrm{d}u$:} \\ &=\frac{1}{2}\int_{\cos 0}^{\cos(\pi^2/16)} -\dfrac{1}{v}\mathrm{d}v \\ \text{\text{Every piece is changed from $u$ to $v$: integrand, differential, limits.}} \\ &=-\frac{1}{2}\int_{1}^{\cos(\pi^2/16)} \dfrac{1}{v}\mathrm{d}v& \cos(0)=1\\ &=-\frac{1}{2}\bigg[\log|v|\bigg]_{1}^{\cos(\pi^2/16)}&\text{FTC Part 2}\\ &=-\frac{1}{2}\left(\log\left(\cos(\pi^2/16)\right)-\log(1)\right)\\ &=-\frac{1}{2}\log\left(\cos(\pi^2/16)\right)& \log(1)=0 \end{aligned}$$

Using the chain rule, we see that $$\frac{\mathrm{d}}{\mathrm{d}x}\{f(g(x))\}=f'(g(x))g'(x)$$ So, $f(g(x))$ is an antiderivative of $f'(g(x))g'(x)$. All antiderivatives of $f'(g(x))g'(x)$ differ by only a constant, so:

$$\begin{aligned} \int f'(g(x))g'(x)\ \mathrm{d}x - f(g(x))&= f(g(x))+C-f(g(x))\\ &=C \end{aligned}$$

That is, our expression simplifies to some constant $C$.

Remark: since $$\int f'(g(x))g'(x)\ \mathrm{d}t - f(g(x))=C$$ we conclude $$\int f'(g(x))g'(x)\ \mathrm{d}t = f(g(x))+C$$ which is precisely how we perform substitution on integrals.

The curves intersect when $y=x$ and $y=x^3-x$. To find these points, we set:

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

For $x\ge 0$, the curves intersect at $(0,0)$ and $(\sqrt{2},\sqrt{2})$.

A handy observation is that, since both curves are continuous and they do not meet each other between $x=0$ and $x=\sqrt{2}$, we don't have to worry about dividing our area into two regions: one of the functions is always on the top, and the other is always on the bottom.

Using vertical strips:

The top and bottom boundaries of the specified region are $y=T(x)=x$ and $y=B(x)=x^3-x$, respectively. So,

$$\begin{aligned} {\rm Area} = \int_0^{\sqrt{2}}\big[T(x)-B(x)\big]\ \mathrm{d}x = \int_0^{\sqrt{2}}\big[x-(x^3-x)\big]\ \mathrm{d}x = \int_0^{\sqrt{2}} 2x-x^3 \ \mathrm{d}x \end{aligned}$$

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.