적분 전체 문제

If $f({t}_{0})>g({t}_{0})$ for some ${t}_{0}\in [a,b],$ then since $f-g$ is continuous, there is an interval containing t₀ such that $f(t)>g(t)$ over the interval $[c,d],$ and then ${\int }_{d}^{d}f(t)dt>{\int }_{c}^{d}g(t)dt$ over this interval.

$\pi -\frac{32}{25}$

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

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, $$\int_{0}^6 (x-3)^3\,\mathrm{d}x = \int_{-3}^3 x^3\,\mathrm{d}x=0$$

(a) $$\begin{aligned} (ax)^2+(by)^2&=1\\ by&=\sqrt{1-(ax)^2}\\ y &= \frac{1}{b}\sqrt{1-(ax)^2} \end{aligned}$$ (b) The values of $x$ in the domain of the function above are those that satisfy $1-(ax)^2 \geq 0$. That is, $-\frac{1}{a}\leq x \leq \frac{1}{a}$. Therefore, the upper half of the ellipse has area

$$\begin{aligned} \displaystyle\frac{1}{b}&\displaystyle\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{1-(ax)^2}\ \mathrm{d}x \\ \text{The upper half of a circle has equation $y=\sqrt{r^2-x^2}$. } \\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{a^2\left(\frac{1}{a^2}-x^2\right)}\ \mathrm{d}x\\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}a\sqrt{\frac{1}{a^2}-x^2}\ \mathrm{d}x\\ &=\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\ \mathrm{d}x \end{aligned}$$ (c) The function $y=\sqrt{\dfrac{1}{a^2}-x^2}$ is the upper-half of the circle centred at the origin with radius $\dfrac{1}{a}$. So, the expression from (b) evaluates to $\left(\dfrac{a}{b}\right)\dfrac{\pi}{2a^2} = \dfrac{\pi}{2ab}$.

The expression from (b) was half of the ellipse, so the area of the ellipse is $\dfrac{\pi}{ab}$.

Remark: this was a slightly long-winded way of getting the result. The reasoning is basically this:

• The area of the unit circle $x^2+y^2=1$ is $\pi$ .
• The ellipse $(ax)^2+y^2=1$ is obtained by shrinking the unit circle horizontally by a factor of $a$. So, its area is $\dfrac{\pi}{a}$ .
• Further, the ellipse $(ax)^2+(by)^2=1$ is obtained from the previous ellipse by shrinking it vertically by a factor of $b$. So, its area is $\dfrac{\pi}{ab}$ .

Let's recall the definitions of even and odd functions: $f(x)$ is even if $f(-x)=f(x)$ for every $x$ in its domain, and $f(x)$ is odd if $f(-x)=-f(x)$ for every $x$ in its domain.

Let $h(x)=f(x)\cdot g(x)$.

• **even $\times$ even: ** If $f$ and $g$ are both even, then $h(-x)=f(-x)\cdot g(-x) = f(x)\cdot g(x)=h(x)$, so their product is even.
• **odd $\times$ odd: ** If $f$ and $g$ are both odd, then $h(-x)=f(-x)\cdot g(-x) =[- f(x)]\cdot [-g(x)]=f(x)\cdot g(x)=h(x)$, so their product is even.
• **even $\times$ odd: ** If $f$ is even and $g$ is odd, then $h(-x)=f(-x)\cdot g(-x) = f(x)\cdot[- g(x)]=-[f(x)\cdot g(x)]=-h(x)$, so their product is odd. Because multiplication is commutative, the order we multiply the functions in doesn't matter.

We note that the table would be the same as if we were adding (not multiplying) even and odd numbers (not functions).

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 $f'(a_i)=f'(-a_i)$.)

The car stops when $v(t)=30-10t=0$, which occurs at time $t = 3$. The distance covered up to that time is

$$\begin{aligned} \int_0^3 v(t)\,\mathrm{d}t = (30t - 5t^2)\Big|_{0}^{3} = (90-45)-0 = 45\,\textrm{m}. \end{aligned}$$