적분 전체 문제

$F(x)=\frac{2}{5}{\left(\sqrt{x}\right)}^{5}+C$

${\int }_{0}^{1}(1-6x+12{x}^{2}-8{x}^{3})dx=(x-3{x}^{2}+4{x}^{3}-2{x}^{4})=(1-3+4-2)(0-0+0-0)=0$

$2\sqrt{x}+C$

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 $f(x)$.

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) False. For example if

$$\begin{aligned} f(x) = \begin{cases} 0 & \text{for $x<0$} \\ 1 & \text{for $x\ge0$} \end{cases} \end{aligned}$$

then $\int_{-3}^{-2} f(x) \mathrm{d}x=0$ and $-\int_{3}^{2} f(x) \mathrm{d}x=\int^{3}_{2} f(x) \mathrm{d}x=1$.

(b) False. For example, if $f(x)=x$, then $\int_{-3}^{-2} f(x)\,\mathrm{d}x$ is negative while $\int_2^3 f(x)\,\mathrm{d}x$ is positive, so they cannot be the same.

(c) False. For example, consider the functions

$$\begin{aligned} f(x) = \begin{cases} 0 & \text{for $x<\frac{1}{2}$} \\ 1 & \text{for $x\ge\frac{1}{2}$} \end{cases} &&\text{and}&&g(x) = \begin{cases} 0 & \text{for $x\ge \frac{1}{2}$} \\ 1 & \text{for $x<\frac{1}{2}$} \end{cases} \end{aligned}$$

Then $f(x)\cdot g(x)=0$ for all $x$, so $\int_0^1 f(x)\cdot g(x) \mathrm{d}x=0$. However, $\int_0^1 f(x) \mathrm{d}x= \frac{1}{2}$ and $\int_0^1 g(x) \mathrm{d}x= \frac{1}{2}$, so $\int_0^1 f(x)\mathrm{d}x \cdot \int_0^1 g(x) \mathrm{d}x= \frac{1}{4}$.

(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.

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