부정적분의 뜻과 계산

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

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

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

$x^6/6+C$

$e^\pi x+C$

$ax+C$

$-\cos x+3$

$5e^x+5$

$x^4-x^3+7$

$\tan x + 4$

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

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

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

$5e^x-2x$

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

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

$3x-2$

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.

Notice $f(x)$ is nonnegative for an interval covering the left part of the graph, and negative on the right part of the graph. That means its antiderivative is increasing for the left interval, then decreasing for the right interval. This applies to $A(x)$ and $C(x)$, but not $B(x)$.

There are only three points where $A(x)$ has a horizontal tangent line: at its global maximum and the endpoints of the interval shown. By contrast, $C(x)$ has a horizontal tangent line in four places: at its global maximum, at its inflection point, and at the endpoints of the interval shown. Since $f(x)=0$ four times (and these line up with the horizontal portions of $C(x)$) we conclude $C(x)$ is the antiderivative of $f(x)$.

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, $$f(x)-\dfrac{\pi}{2}=\arcsin\left(\sin\left[f(x)-\dfrac{\pi}{2}\right]\right)=\arcsin(2x+1)=g(x).$$