적분 전체 문제

$16$

$1/(e-1)$

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

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

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

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

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

Using the identity

$$\begin{aligned} \int\limits_a^b f(x)\ \mathrm{d}x &= \int\limits_a^c f(x)\ \mathrm{d}x+ \int\limits_c^b f(x)\ \mathrm{d}x\ , \\ \text{ we see} \\ \int\limits_a^b\cos x\ \mathrm{d}x &= \int\limits_a^0 \cos x\ \mathrm{d}x+ \int\limits_0^b \cos x\ \mathrm{d}x\\&= -\int\limits_0^a \cos x\ \mathrm{d}x+ \int\limits_0^b \cos x\ \mathrm{d}x\\ &=-\sin a + \sin b\\ &=\sin b - \sin a \end{aligned}$$

(a) $\Delta x = \dfrac{b-a}{n}=\dfrac{0-5}{100} = -\dfrac{1}{20}$

Note: if we were to use the Riemann-sum definition of a definite integral, this is how we would justify the identity $\int\limits_a^b f(x)\mathrm{d}x=-\int\limits_b^a f(x)\mathrm{d}x$. (b) The heights of the rectangles are given by $f(x_i)$, where $x_i = a+i\Delta x = 5 - \frac{i}{20}$. Since $f(x)$ only gives positive values, $f(x_i) >0$, so the heights of the rectangles are positive. (c) Our Riemann sum is the sum of the signed areas of individual rectangles. Each rectangle has a negative base ($\Delta x$) and a positive height ($f(x_i)$). So, each term of our sum is negative. If we add up negative numbers, the sum is negative. So, the Riemann sum is negative. (d) Since $f(x)$ is always above the $x$-axis, $\int\limits_0^5 f(x)\mathrm{d}x$ is positive.

We first use additivity:

$$\begin{aligned} \int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\mathrm{d}x = \int_{-2}^{2} 5\,\mathrm{d}x + \int_{-2}^{2} \sqrt{4-x^2}\,\mathrm{d}x \end{aligned}$$

The first integral represents the area of a rectangle of height 5 and width 4 and so equals $20$. The second integral represents the area above the $x$–axis and below the curve $y=\sqrt{4-x^2}$ or $x^2+y^2=4$. That is a semicircle of radius 2, which has area $\frac{1}{2}\pi 2^2$. So

$$\begin{aligned} \int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\mathrm{d}x = 20 + 2\pi \end{aligned}$$

Our integrand $f(x)=(x-3)^3$ is neither even nor odd. However, it does have a similar symmetry. Namely, $f(3+x)=-f(3-x)$. So, $f$ is “negatively symmetric" across the line $x=3$. This suggests that the integral should be 0: the positive area to the right of $x=3$ will be the same as the negative area to the left of $x=3$.

Another way to see this is to notice that the graph of $f(x)=(x-3)^3$ is equivalent to the graph of $g(x)=x^3$ shifted three units to the right, and $g(x)$ is an odd function. So, $$\int_{0}^6 (x-3)^3\,\mathrm{d}x = \int_{-3}^3 x^3\,\mathrm{d}x=0$$

Since $f(x)$ is odd, $f(0)=-f(-0)=-f(0)$. So, $f(0)=0$.

However, this restriction does not apply to $g(x)$. For example, for any constant $c$, let $g(x)=c$. Then $g(x)$ is even and $g(0)=c$. So, $g(0)$ can be any real number.

Let $x$ be any real number.

• $f(x)=f(-x)$ (since $f(x)$ is even), and
• $f(x)=-f(-x)$ (since $f(x)$ is odd).
• So, $f(x)=-f(x)$.
• Then (adding $f(x)$ to both sides) we see $2f(x)=0$, so $f(x)=0$.

So, $f(x)=0$ for every $x$.

• Solution 1: Suppose $f(x)$ is an odd function. We investigate $f'(x)$ using the chain rule:

$$\begin{aligned} f(-x)&=-f(x)& \text{(odd function)}\\ \frac{\mathrm{d}}{\mathrm{d}x}\{f(-x)\}&=\frac{\mathrm{d}}{\mathrm{d}x}\{-f(x)\}\\ -f'(-x)&=-f'(x) & \text{(chain rule)}\\ f'(-x)&=f'(x) \end{aligned}$$

So, when $f(x)$ is odd, $f'(x)$ is even.

Similarly, suppose $f(x)$ is even.

$$\begin{aligned} f(-x)&=f(x)& \text{(even function)}\\ \frac{\mathrm{d}}{\mathrm{d}x}\{f(-x)\}&=\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\\ -f'(-x)&=f'(x) & \text{(chain rule)}\\ f'(-x)&=-f'(x) \end{aligned}$$

So, when $f(x)$ is even, $f'(x)$ is odd.

• Solution 2: Another way to think about this problem is to notice that “mirroring" a function changes the sign of its derivative. Then since an even function is “mirrored once" (across the $y$-axis), it should have $f'(x)=-f'(-x)$, and so the derivative of an even function should be an odd function. Since an odd function is “mirrored twice" (across the $y$-axis and across the $x$-axis), it should have $f'(x)=-(-f'(-x))=f'(-x)$. So the derivative of an odd function should be even. These ideas are presented in more detail below.

First, we consider the case where $f(x)$ is even, and investigate $f'(x)$.

The whole function has a mirror-like symmetry across the $y$-axis. So, at $x$ and $-x$, the function will have the same “steepness," but if one is increasing then the other is decreasing. That is, $f'(-x)=-f'(x)$. (In the picture above, compare the slope at some point $a_i$ with its corresponding point $-a_i$.) So, $f'(x)$ is odd when $f(x)$ is even.

Second, let's consider the case where $f(x)$ is odd, and investigate $f'(x)$. Suppose the blue graph below is $y=f(x)$. If $f(x)$ were even, then to the left of the $y$-axis, it would look like the orange graph, which we'll call $y=g(x)$.

From our work above, we know that, for every $x>0$, $-f'(x)=g'(-x)$. When $x<0$, $f(x)=-g(x)$. So, if $x>0$, then $-f'(x)=g'(-x)=-f'(-x)$. In other words, $f'(x)=f'(-x)$. Similarly, if $x<0$, then $f'(x)=-g'(x)=f'(-x)$. Therefore $f'(x)$ is even. (In the graph below, you can anecdotally verify that $f'(a_i)=f'(-a_i)$.)

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 $\sin (x^2)$.