부정적분의 뜻과 계산

$3^t/\ln 3+C$

$\frac{5^t}{2\ln 5}+C$

$4/3t^3+6t^2+9t+C$

$t^6/6+t^4/4-3t^2+C$

$x^6/6+C$

$e^\pi x+C$

$ax+C$

$-\cos x+3$

$5e^x+5$

$x^4-x^3+7$

$\tan x + 4$

$7^x/\ln 7 + 1-49/\ln 7$

$5/2x^2+7x+3$

$\frac{7 x^3}{6}-\frac{9 x}{2}+\frac{40}{3}$

$5e^x-2x$

$\theta-\sin (\theta)-\pi +4$

$2x^4+\cos x+\frac{2^x}{(\ln 2)^2}+(5-\frac1{\ln 2})x -1 - \frac{1}{(\ln 2)^2}$

$3x-2$

An antiderivative of $f'(x)$ is a function whose derivative is $f'(x)$. Our original function $f(x)$ has this property, so $f(x)$ is an antiderivative of $f'(x)$, but it's not the most general. We can add a constant to $f(x)$ without affecting its derivative. The most general antiderivative of $f'(x)$ is $f(x)+C$, where $C$ is any constant.

We differentiate $f(x)$ and $g(x)$ using the chain rule. Recall $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin(x)\right\}=\dfrac{1}{\sqrt{1-x^2}}$.

$$\begin{aligned} f'(x)&=\frac{2}{\sqrt{1- \sqrt{x}^2}}\cdot\frac{\mathrm{d}}{\mathrm{d}x} \sqrt{x}\\ &=\frac{2}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}\\ &=\frac{1}{\sqrt{x-x^2}}\\ g'(x)&=\frac{1}{\sqrt{1- (2x-1)^2}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ 2x-1\}\\ &=\frac{2}{\sqrt{1-4x^2+4x-1}}\\ &=\frac{1}{\sqrt{x-x^2}}\\ \end{aligned}$$

So, the derivative of $f(x)-g(x)$ is $0$, which implies that $f(x)$ and $g(x)$ differ by a constant–perhaps a surprising result!

Remark: we don't need calculus to show that $f(x)$ and $g(x)$ only differ by a constant. Define $\theta=\sin^{-1}\sqrt{x}$, so that $\sin\theta=\sqrt{x}$ and $f(x) = 2\theta$ , and then $$\sin \left[f(x)-\frac{\pi}{2}\right]=-\cos f(x) =-\cos 2\theta=-[1-2\sin^2\theta] =2\sin^2\theta -1=2x-1$$ Our goal is to take the arcsine of the first and last expressions and conclude $f(x)-\dfrac{\pi}{2}=\arcsin(2x-1)=g(x)$. However, before we can say $\arcsin\left(\sin\left[f(x)-\dfrac{\pi}{2}\right]\right)=f(x)-\dfrac{\pi}{2}$, we have to check that $-\dfrac{\pi}{2} \leq f(x)-\dfrac{\pi}{2}\leq\dfrac{\pi}{2}$. That is, we need to show that $0 \leq f(x) \leq \pi$.

As $0\le\theta\le\frac{\pi}{2}$ (since $\sin^{-1}$ always takes values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ and since $\sqrt{x}\ge 0$), we have that $0\le f(x)=2\theta\le \pi$. So, $$f(x)-\dfrac{\pi}{2}=\arcsin\left(\sin\left[f(x)-\dfrac{\pi}{2}\right]\right)=\arcsin(2x+1)=g(x).$$