미적분의 기본정리

$2$

$16/3$

$16$

$1/(e-1)$

$F'(x) = (3x^2+1)\frac{1}{x^3+x}$

$F'(x) = -3x^{11}$

$F'(x) = 2x(x^2+2)-(x+2)$

$F'(x) = e^x\sin (e^x) - 1/x\sin(\ln x)$

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

“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 $\displaystyle\int_{-2}^0 \frac{1}{x+1}\ \mathrm{d}x$.

By the Fundamental Theorem of Calculus,

$$\begin{aligned} \int_0^2 \big(x^3+\sin x)\,\mathrm{d}x &=\left[\frac{x^4}{4}-\cos x\right]_0^2\\ &=\left(\frac{2^4}{4}-\cos 2\right)-\left(0 -\cos 0\right) \\ &= 4-\cos2+1=5-\cos2. \end{aligned}$$

The integrand is similar to $\dfrac{1}{1+x^2}$, which is the derivative of arctangent. Indeed, we have

$$\begin{aligned}\displaystyle\int \dfrac{1}{1+25x^2}\mathrm{d}x &= \int \dfrac{1}{1+(5x)^2}\mathrm{d}x. \\ \text{ So, a reasonable first guess for the antiderivative might be } \\ F(x) &\stackrel{?}{=} \arctan(5x). \\ \text{ However, because of the chain rule, } \\ F'(x) &= \dfrac{5}{1+(5x)^2}. \\ \text{ In order to ``fix" the numerator, we make a second guess: } \\ F(x) &= \frac{1}{5}\arctan(5x)\\ F'(x) &= \dfrac{1}{5}\left(\dfrac{5}{1+(5x)^2}\right) = \dfrac{1}{1+25x^2}\\ \text{So,}\qquad \displaystyle\int \dfrac{1}{1+25x^2}\mathrm{d}x&=\frac{1}{5}\arctan(5x)+C. \end{aligned}$$

The integrand is similar to $\dfrac{1}{\sqrt{1-x^2}}$. In order to formulate a guess for the antiderivative, let's factor out $\sqrt{2}$ from the denominator:

$$\begin{aligned} \displaystyle\int \dfrac{1}{\sqrt{2-x^2}}\mathrm{d}x&= \displaystyle\int \dfrac{1}{\sqrt{2\left(1-\frac{x^2}{2}\right)}}\mathrm{d}x\\ &=\displaystyle\int \dfrac{1}{ \sqrt{2}\sqrt{1-\frac{x^2}{2}}}\mathrm{d}x\\ &= \int\frac{1}{\sqrt{2}}\cdot \dfrac{1}{ \sqrt{1-\left(\dfrac{x}{\sqrt2}\right)^2}}\mathrm{d}x\\ \\ \text{At this point, we might guess that our antiderivative is something like $F(x) = \arcsin\left(\dfrac{x}{\sqrt{2}}\right)$. To explore this possibility, we can differentiate, and see what we get.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin\left(\dfrac{x}{\sqrt{2}}\right)\right\}&=\frac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{1-\left(\dfrac{x}{\sqrt{2}}\right)^2}} \\ \text{This is exactly what we want! So,} \\ \int \dfrac{1}{\sqrt{2-x^2}}\mathrm{d}x&=\arcsin\left(\dfrac{x}{\sqrt2}\right)+C \end{aligned}$$

We know that $\int \sec^2 x\ \mathrm{d}x = \tan x +C$, and $\sec^2x = \tan^2 x + 1$, so

$$\begin{aligned} \int \tan^2 x\ \mathrm{d}x &= \int \sec^2x - 1 \ \mathrm{d}x\\ &=\int \sec^2 x \ \mathrm{d}x - \int 1 \ \mathrm{d}x\\ &=\tan x - x + C \end{aligned}$$

• Solution 1: This might not obviously look like the derivative of anything familiar, but it does look like half of a familiar trig identity: $2\sin x \cos x = \sin(2x)$.

$$\begin{aligned} \int 3 \sin x \cos x \ \mathrm{d}x&=\int \frac{3}{2}\cdot 2\sin x \cos x \ \mathrm{d}x\\ &=\int \frac{3}{2} \sin(2x)\ \mathrm{d}x \\ \text{So, we might guess that the antiderivative is something like $-\cos(2x)$. We only need to figure out the constants.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{-\cos(2x)\}&=2\sin(2x)\\ \text{So,}\qquad \frac{\mathrm{d}}{\mathrm{d}x}\left\{-\frac{3}{4}\cos(2x)\right\}&=\frac{3}{2}\sin(2x)\\ \text{Therefore,}\qquad \int 3\sin x \cos x\ \mathrm{d}x&=-\frac{3}{4}\cos(2x)+C \end{aligned}$$

• Solution 2: You might notice that the integrand looks like it came from the chain rule, since $\cos x$ is the derivative of $\sin x$. Using this observation, we can work out the antideriative:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sin^2 x\right\}&=2\sin x \cos x\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{3}{2}\sin^2 x\right\}&=3\sin x \cos x\\ \text{So,}\qquad \int 3\sin x \cos x \ \mathrm{d}x&=\frac{3}{2}\sin^2 x+C \end{aligned}$$

These two answers look different. Using the identity $\cos(2x)=1 - 2 \sin^2(x)$, we reconcile them:

$$\begin{aligned} -\frac{3}{4}\cos(2x)+C&= -\frac{3}{4}\left(1-2\sin^2x\right)+C\\ &=\frac{3}{2}\sin^2 x + \left(C-\frac{3}{4}\right) \end{aligned}$$

The $\frac{3}{4}$ here is not significant. Remember that $C$ is used to designate a constant that can take any value between $-\infty$ and $+\infty$. So $C-\frac{3}{4}$ is also just a constant that can take any value between $-\infty$ and $+\infty$. As the two answers we found differ by a constant, they are equivalent.

It's not immediately obvious which function has $\cos^2 x$ as its derivative, but we can make the situation a little clearer by using the identity $\cos^2 x = \dfrac{1+\cos(2x)}{2}$:

$$\begin{aligned} \int \cos^2 x \ \mathrm{d}x&=\int \frac{1}{2}\cdot \left(1+\cos(2x)\right)\ \mathrm{d}x\\ &=\int \frac{1}{2}\ \mathrm{d}x + \int \frac{1}{2}\cos(2x)\ \mathrm{d}x\\ &=\frac{1}{2}x + C+ \int \frac{1}{2}\cos(2x)\ \mathrm{d}x \\ \text{For the remaining integral, we might guess something like $F(x) = \sin(2x)$. Let's figure out the appropriate constant:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sin(2x) \right\}&=2\cos(2x)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{1}{4}\sin(2x) \right\}&=\frac{1}{2}\cos(2x)\\ \text{So,}\qquad \int \frac{1}{2}\cos(2x)\ \mathrm{d}x &=\frac{1}{4}\sin(2x)+C\\ \text{Therefore,}\qquad \int \cos^2 x \ \mathrm{d}x&=\frac{1}{2}x+\frac{1}{4}\sin(2x)+C \end{aligned}$$

By the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} F'(x)&=\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x \log(2+\sin t)\,\mathrm{d}t &&=\log(2+\sin x) \\ G'(y)&=\frac{\mathrm{d}}{\mathrm{d}y}\bigg[- \int_0^y \log(2+\sin t)\,\mathrm{d}t\bigg] &&=-\log(2+\sin y) \\ \end{aligned}$$

So,

$$F'\Big(\frac{\pi}{2}\Big)=\log 3\qquad G'\Big(\frac{\pi}{2}\Big)=-\log(3)$$