미적분의 기본정리

First, let's find a general antiderivative of $x^3-\sin(2x)$.

• One function with derivative $x^3$ is $\dfrac{x^4}{4}$.
• To find an antiderivative of $\sin(2x)$, we might first guess $\cos(2x)$; checking, we see $\frac{\mathrm{d}}{\mathrm{d}x}\{\cos(2x)\}=-2\sin(2x)$. So, we only need to multiply by $-\dfrac{1}{2}$: $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\left\{-\dfrac{1}{2}\cos 2x\right\}=\sin(2x)$.

So, the general antiderivative of $f(x)$ is $\dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+C$. To satisfy $F(0)=1$, we need (The symbol $\iff$ is read “if and only if”. This is used in mathematics to express the logical equivalence of two statements. To be more precise, the statement $P \iff Q$ tells us that $P$ is true whenever $Q$ is true and $Q$ is true whenever $P$ is true.)

$$\begin{aligned} \Big[\frac{x^4}{4}+\frac{1}{2}\cos 2x+C\Big]_{x=0}=1 \iff \frac{1}{2} + C = 1 \iff C=\frac{1}{2} \end{aligned}$$

So $F(x) = \dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+\dfrac{1}{2}$.

This is a tempting thought:

$$\begin{aligned} \int \frac{1}{x}\ \mathrm{d}x&=\log|x|+C \\ \text{ so perhaps similarly } \\ \int \frac{1}{x^2}\ \mathrm{d}x&\stackrel{?}{=}\log|x^2|+C=\log(x^2)+C \\ \text{ We check by differentiating:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{\log(x^2)\} &= \frac{\mathrm{d}}{\mathrm{d}x}\{2\log x\}=\frac{2}{x} \neq \frac{1}{x^2} \end{aligned}$$ So, it wasn't so easy: false.

When we're guessing antiderivatives, we often need to adjust our original guesses a little. Changing constants works well; changing functions usually does not.

This is tempting:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\sin(e^x)\} &= e^x\cos(e^x) \\ \text{so perhaps} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{\sin(e^x)}{e^x}\right\} &\stackrel{?}{=} \cos(e^x) \\ \text{We check by differentiating:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{\sin(e^x)}{e^x}\right\} &=\frac{e^x\left(\cos(e^x)\cdot e^x\right)-\sin(e^x)e^x}{e^{2x}} &\text{(quotient rule)}\\ & = \cos (e^x) - \frac{\sin(e^x)}{e^x}\\ &\neq \cos(e^x) \end{aligned}$$

So, the statement is false.

When we're guessing antiderivatives, we often need to adjust our original guesses a little. Dividing by constants works well; dividing by functions usually does not.

“The instantaneous rate of change of $F(x)$ with respect to $x$" is another way of saying “$F'(x)$". From the Fundamental Theorem of Calculus Part 1, we know this is $\sin (x^2)$.

The slope of the tangent line to $y=F(x)$ when $x=3$ is exactly $F'(3)$. By the Fundamental Theorem of Calculus Part 1, $F'(x) = e^{1/x}$. Then $F'(3) = e^{1/3} = \sqrt[3]{e}$.

For any constant $C$, $F(x)+C$ is an antiderivative of $f(x)$, because $\frac{\mathrm{d}}{\mathrm{d}x}\{F(x)+C\} = \frac{\mathrm{d}}{\mathrm{d}x}\{F(x)\} = f(x)$. So, for example, $F(x)$ and $F(x)+1$ are both antiderivatives of $f(x)$.

(a) The antiderivative of $\cos x$ is $\sin x$, and $\cos x$ is continuous everywhere, so $\displaystyle\int_{-\pi}^\pi \cos x \ \mathrm{d}x = \sin(\pi)-\sin(-\pi) = 0$. (b) Since $\sec^2 x$ is discontinuous at $x=\pm\frac{\pi}{2}$, the Fundamental Theorem of Calculus Part 2 does not apply to $\displaystyle\int_{-\pi}^\pi \sec^2 x \ \mathrm{d}x$. (c) Since $\frac{1}{x+1}$ is discontinuous at $x=-1$, the Fundamental Theorem of Calculus Part 2 does not apply to $\displaystyle\int_{-2}^0 \frac{1}{x+1}\ \mathrm{d}x$.