적분법 전체 문제

Let A be the area of the unit circle. The circle encloses n congruent triangles each of area $\frac{\sin \left(\frac{2\pi }{n}\right)}{2},$ so $\frac{n}{2}\sin \left(\frac{2\pi }{n}\right)\le A.$ Similarly, the circle is contained inside n congruent triangles each of area $\frac{BH}{2}=\frac{1}{2}{\left( \cos \left( \frac{\pi }{n} \right)+\sin \left( \frac{\pi }{n} \right)\tan \left( \frac{\pi }{n} \right) \right)}^{2}\sin \left( \frac{2\pi }{n} \right),$ so $A\le \frac{n}{2}\sin \left(\frac{2\pi }{n}\right)(\cos \left(\frac{\pi }{n}\right)+\sin \left(\frac{\pi }{n}\right)\tan {\left(\frac{\pi }{n}\right))}^{2}.$ As $n\to \infty ,\frac{n}{2}\sin \left(\frac{2\pi }{n}\right)=\frac{\sin \left(\frac{2\pi }{n}\right)}{\left(\frac{2}{n}\right)}A\to \pi ,$ so we conclude $\pi \le A.$ Also, as $n\to \infty ,\cos \left(\frac{\pi }{n}\right)+\sin \left(\frac{\pi }{n}\right)\tan \left(\frac{\pi }{n}\right)\to 1,$ so we also have $A\le \pi .$ By the squeeze theorem for limits, we conclude that $A=\pi .$

${L}_{N}=\frac{b-a}{N}\sum_{i=1}^{N}f\left(a+(b-a)\frac{i-1}{N}\right)=\frac{b-a}{N}\sum_{i=0}^{N-1}f\left(a+(b-a)\frac{i}{N}\right)$ and ${R}_{N}=\frac{b-a}{N}\sum_{i=1}^{N}f\left(a+(b-a)\frac{i}{N}\right).$ The left sum has a term corresponding to $i=0$ and the right sum has a term corresponding to $i=N.$ In ${R}_{N}-{L}_{N},$ any term corresponding to $i=1,2,\ldots ,\,N-1$ occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with ${R}_{N}-{L}_{N}=\frac{b-a}{N}\left(f(a+(b-a))\frac{N}{N}\right)-\left(f(a)+(b-a)\frac{0}{N}\right)=\frac{b-a}{N}(f(b)-f(a)).$

$u=\sin ({t}^{2});$ the integral becomes $\frac{1}{2}{\int }_{0}^{0}udu.$

$u=\left(1+{\left(t-\frac{1}{2}\right)}^{2}\right);$ the integral becomes $-{\int }_{5/4}^{5/4}\frac{1}{u}du.$

$u=1-t;$ the integral becomes $\begin{array}{c}{\int }_{1}^{-1}u\cos (\pi (1-u))du \\ ={\int }_{1}^{-1}u[\cos \pi \cos \pi u-\sin \pi \sin \pi u]du \\ =-{\int }_{1}^{-1}u\cos \pi udu \\ ={\int }_{-1}^{1}u\cos \pi udu=0\end{array}$ since the integrand is odd.

${\int }_{a}^{b}\frac{dx}{x}=\ln (b)-\ln (a)=\ln \left(\frac{1}{a}\right)-\ln \left(\frac{1}{b}\right)={\int }_{1/b}^{1/a}\frac{dx}{x}$

Answers will vary.

$x=E(\ln (x)).$ Then, $1=\frac{E'(\ln x)}{x}\,\text{or}\,x=E'(\ln x).$ Since any number t can be written $t=\ln x$ for some x, and for such t we have $x=E(t),$ it follows that for any $t,E'(t)=E(t).$

Answers may vary

All the antiderivatives differ only by a constant, so we can write them all as $v(x)+C$ for some $C$. Then, using the formula for integration by parts,

$$\begin{aligned} \int u(x)\cdot v'(x) \mathrm{d}x&=\underbrace{u(x)}_u\underbrace{\big[ v(x)+C\big]}_v - \int \underbrace{\big[ v(x)+C\big]}_v \underbrace{u'(x)\mathrm{d}x}_{\mathrm{d}u}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\mathrm{d}x - \int Cu'(x)\mathrm{d}x \\&=u(x)v(x)+Cu(x) - \int v(x)u'(x)\mathrm{d}x - Cu(x)+D \\&=u(x)v(x) - \int v(x)u'(x)\mathrm{d}x +D \end{aligned}$$

where $D$ is any constant.

Since the terms with $C$ cancel out, it didn't matter what we chose for $C$–all choices end up the same.

(a) The average value of $f$ on the interval $[0,\sqrt{2}]$ is $$\begin{aligned}r &= \frac{1}{\sqrt{2}} \int_0^{\sqrt{2}} 2-x^2 \, dx \\ &= \left. \frac{1}{\sqrt{2}}\left(2x - \frac{x^3}{3}\right) \right|_0^{\sqrt{2}} \\ &= \frac{1}{\sqrt{2}}\left( 2\sqrt{2} - \frac{\sqrt{2}^3}{3} \right) \\ &= 2 - \frac{2}{3} \\ &= \frac{4}{3}\end{aligned}\,.$$ (b) The intersection point of $y=f(x)$ and $y=r$ occurs, for positive $x$, when $2-x^2 = \frac{4}{3}$, from which it follows that $x^2 = \frac{2}{3}$ and thus $x = \frac{\sqrt{6}}{3}$. A graph of $f$, $y = r$, and the intersection point is shown in the following figure. (c) Let $A_1$ be the area that lies below $y = f(x)$ and above $y = r$ on the interval $\left[0, \frac{\sqrt{6}}{3}\right]$ and $A_2$ the are of the region that lies below $y = r$ and above $y = f(x)$ on $\left[\frac{\sqrt{6}}{3}, \sqrt{2}\right]$ as illustrated in the figure below. Then $$\begin{aligned}A_1 &= \int_0^{\sqrt{6}/3} (2-x^2) - \frac{4}{3} \ dx \\ &= \int_0^{\sqrt{6}/3} \frac{2}{3} - x^2 \ dx \\ &= \left. \left( \frac{2}{3}x - \frac{x^3}{3} \right) \right|_0^{\sqrt{6}/3} \\ &= \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) - \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 \\ &= \frac{2\sqrt{6}}{9} - \frac{6 \sqrt{6}}{81} \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\,.$$ and $$\begin{aligned}A_2 &= \int_{\sqrt{6}/3}^{\sqrt{2}} \frac{4}{3} - (2-x^2) \ dx \\ &= \int_{\sqrt{6}/3}^{\sqrt{2}} x^2 - \frac{2}{3} \ dx \\ &= \left. \left(\frac{x^3}{3} - \frac{2}{3}x \right) \right|_{\sqrt{6}/3}^{\sqrt{2}} \\ &= \left(\frac{2\sqrt{2}}{3} - \frac{2}{3} \sqrt{2} \right) - \left[ \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 - \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) \right] \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\,.$$ Hence $A_1 = A_2$ as desired. (d) The answer is yes. Let $AV$ is the average value of a function $f$ on an interval $[a,b]$. By the definition of average value, $$AV(b-a) = \int_a^b f(x) \ dx\,,$$ which can also be viewed as the area of the rectangle with base $(b-a)$ and height $AV$. Now suppose that the area of the region above the graph of $f$ and below the line $y=AV$ on the interval $[a,b]$ is $R_1$ and the area of the region below the graph of $f$ and above the line $y=AV$ on the interval $[a,b]$ is $R_2$. The figure below illustrates that the value of the definite integral is the area of the rectangle plus the area above the rectangle minus the area below the rectangle, or $$\int_a^b f(x) \, dx = AV(b-a) + R_2 - R_1\,.$$ Having already established that $AV(b-a) = \int_a^b f(x) \ dx$, it follows that $R_2 - R_1 = 0$. So the area of the region that lies below $y = f(x)$ and above $y = AV$ is equal to the amount of area that lies below $y = AV$ and above $y = f(x)$ on $[a,b]$.