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.

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.

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