적분 기법 전체 문제

If $b= \pm \infty$, then our integral is improper because one limit is not a real number.

Furthermore, our integral will be improper if its domain of integration contains either of its infinite discontinuities, $x=1$ and $x=-1$. Since one limit of integration is 0, the integral is improper if $b \geq 1$ or if $b \leq -1$.

Below, we've graphed $\frac{1}{x^2-1}$ to make it clearer why values of $b$ in $(-1,1)$ are the only values that don't result in an improper integral when the other limit of integration is $a=0$.

Since the integrand is continuous for all real $x$, the only kind of impropriety available to us is to set $b = \pm\infty$.

The denominator is zero when $x=1$, but the numerator is not, so the integrand has a singularity (infinite discontinuity) at $x=1$. Let's replace the limit $x=1$ with a variable that creeps toward 1.

$$\int_0^1\frac{x^4}{x^5-1}\,\mathrm{d}x =\lim_{t\rightarrow 1^-}\int_0^t\frac{x^4}{x^5-1}\,\mathrm{d}x$$

To evaluate this integral we use the substitution $u=x^5$, $\mathrm{d}u=5x^4\mathrm{d}x$. When $x=0$ we have $u=0$, and when $x=t$ we have $u=t^5$, so

$$\begin{aligned} \int_0^1\frac{x^4}{x^5-1}\,\mathrm{d}x&=\lim_{t \to 1^-}\int_{0}^{t}\frac{x^4}{x^5-1}\,\mathrm{d}x =\lim_{t \to 1^-}\int_{u=0}^{u=t^5}\frac{1}{5(u-1)}\,\mathrm{d}u\\ &=\lim_{t \to 1^-}\left( \left[\frac{1}{5}\log|u-1|\right]_0^{t^5}\right) =\lim_{t \to 1^-}\frac{1}{5}\log|t^5-1|=-\infty \end{aligned}$$

The limit does not exist, so the integral diverges.

The denominator of the integrand is zero when $x=-1$, but the numerator is not. So, the integrand has a singularity (infinite discontinuity) at $x=-1$. This is the only “source of impropriety" in this integral, so we only need to make one break in the domain of integration.

$$\int_{-2}^2\frac{1}{(x+1)^{4/3}}\,\mathrm{d}x =\lim_{t\rightarrow -1^-}\int_{-2}^t \frac{1}{(x+1)^{4/3}}\,\mathrm{d}x +\lim_{t\rightarrow -1^+}\int_t^2 \frac{1}{(x+1)^{4/3}}\,\mathrm{d}x$$

Let's start by considering the left limit.

$$\begin{aligned} \lim_{t \to -1^-}\int_{-2}^t \frac{1}{(x+1)^{4/3}}\,\mathrm{d}x &=\lim_{t \to -1^-}\left(\left[-\frac{3}{(x+1)^{1/3}}\right|_{-2}^t\right)\\ &=\lim_{t \to -1^-}\left(-\frac{3}{(t+1)^{1/3}}+\frac{3}{(-1)^{1/3}}\right)=\infty \end{aligned}$$

Since this limit diverges, the integral diverges. (A similar argument shows that the second integral diverges. Either one of them diverging is enough to conclude that the original integral diverges.)

There are two “sources of impropriety": the two (infinite) limits of integration. So, we break our integral into two pieces.

$$\begin{aligned} \int_{-\infty}^\infty \cos x~\mathrm{d}x&=\int_{-\infty}^0 \cos x~\mathrm{d}x+\int_{0}^\infty \cos x~\mathrm{d}x\\ &=\lim_{a \to \infty }\left[\int_{-a}^0 \cos x~\mathrm{d}x\right]+ \lim_{b \to \infty }\left[\int_{0}^b \cos x~\mathrm{d}x\right] \\ \text{These are easy enough to antdifferentiate.} \\ &=\lim_{a \to \infty }\left[\sin 0 - \sin(- a) \right]+ \lim_{b \to \infty }\left[\sin b - \sin 0 \right]\\ &=\text{DNE} \end{aligned}$$

Since the limits don't exist, the integral diverges. (It happens that both limits don't exist; even if only one failed to exist, the integral would still diverge.)

To find the volume of the solid, we cut it into horizontal slices, which are thin circular disks. At height $y$, the disk has radius $x=\frac{1}{y}$ and thickness $\mathrm{d}y$, so its volume is $\frac{\pi}{y^2}\mathrm{d}y$. The base of the solid is at height $y=1$, and its top is at height $y=\frac{1}{a}$. So, the volume of the entire solid is:

$$\begin{aligned} \int_1^{1/a} \frac{\pi}{y^2}\mathrm{d}y = \left[-\frac{\pi}{y}\right]_1^{1/a} = \pi (1-a) \end{aligned}$$

If we imagine sliding $a$ closer and closer to 0, the volume increases, getting closer and closer to $\pi$ units, but never quite reaching it.

So, the statement is false. For example, if we set $M=4$, no matter which $a$ we choose our solid has volume strictly less than $M$.

Remark: we've seen before that $\int_0^1 \frac{1}{x}\,\mathrm{d}x$ diverges. If we imagine the solid that would result from choosing $a=0$, it would have a scant volume of $\pi$ cubic units, but a silhouette (side view) of infinite area.

There are two sources of error: the upper bound is $t$, rather than infinity, and we're using an approximation with some finite number of intervals, $n$. Our plan is to first find a value of $t$ that introduces an error of no more than $\frac{1}{2}10^{-4}$. That is, we'll find a value of $t$ such that $\int_t^\infty \frac{e^{-x}}{1+x}~\mathrm{d}x \leq \frac{1}{2}10^{-4}$. After that, we'll find a value of $n$ that approximates $\int_0^t \frac{e^{-x}}{x+1}~\mathrm{d}x$ to within $\frac{1}{2}10^{-4}$. Then, all together, our error will be at most $\frac{1}{2}10^{-4} + \frac{1}{2}10^{-4} = 10^{-4}$, as desired. (Note we could have broken up the error in another way—it didn't have to be $\frac{1}{2}10^{-4}$ and $\frac{1}{2}10^{-4}$. This will give us one of many possible answers.)

Let's find a $t$ such that $\int_t^\infty \frac{e^{-x}}{1+x}\ \mathrm{d}x\le\frac{1}{2} 10^{-4}$. For all $x\ge 0$, $0<\frac{e^{-x}}{1+x}\le e^{-x}$, so

$$\begin{aligned} \int_t^\infty \frac{e^{-x}}{1+x}\ \mathrm{d}x \le \int_t^\infty e^{-x}\ \mathrm{d}x =e^{-t}&\stackrel{(*)}{\le}\frac{1}{2} 10^{-4}\\ \text{where ($*$) is true if }\quad t&\ge -\log\Big(\frac{1}{2} 10^{-4}\Big)\approx 9.90 \end{aligned}$$

Choose, for example, $t=10$.

Now it's time to decide how many intervals we're going to use to approximate $\displaystyle\int_0^t \frac{e^{-x}}{x+1}~\mathrm{d}x$. Again, we want our error to be less than $\frac{1}{2}10^{-4}$. To bound our error, we need to know the second derivative of $\frac{e^{-x}}{x+1}$.

$$\begin{aligned} f(x)=\frac{e^{-x}}{1+x} \implies f'(x)=-\frac{e^{-x}}{1+x}-\frac{e^{-x}}{(1+x)^2} \implies f''(x)&=\frac{e^{-x}}{1+x}+2\frac{e^{-x}}{(1+x)^2} +2\frac{e^{-x}}{(1+x)^3} \end{aligned}$$

Since $f''(x)$ is positive, and decreases as $x$ increases,

$$\begin{aligned} |f''(x)|\le f''(0) = 5 \implies |E_n|\le\frac{5(10-0)^3}{24n^2}=\frac{5000}{24n^2} =\frac{625}{3n^2} \end{aligned}$$

and $|E_n|\le \frac{1}{2} 10^{-4}$ if

$$\begin{aligned} &&\frac{625}{3n^2}&\le \frac{1}{2} 10^{-4}\\ &\iff\qquad & n^2&\ge \frac{1250\times 10^4}{3}\\ &\iff\quad &n&\ge \sqrt{\frac{1.25\times 10^7}{3}} \approx 2041.2 \end{aligned}$$

So $t=10$ and $n= 2042$ will do the job. There are many other correct answers.

Define $F(x)= \int_0^x \frac{1}{e^{t}}\mathrm{d}t$.

$$\begin{aligned} F(x)=\int_0^x \frac{1}{e^{t}}\mathrm{d}t&=\left[-\frac{1}{e^{t}}\right]_0^x = \frac{1}{e^0}-\frac{1}{e^{x}}<\frac{1}{e^0}=1 \end{aligned}$$

So, the statement is false: there is no $x$ such that $F(x)=1$. For every real $x$, $F(x) < \frac{1}{e^0}=1$.

We note here that $\displaystyle\lim_{x \to \infty} \int_0^x\frac{1}{e^{t}}\mathrm{d}t=1$. So, as $x$ grows larger, the gap between $F(x)$ and 1 grows infintesimally small. But there is no real value of $x$ where $F(x)$ is exactly equal to 1.

First, we note the integral is improper. So, we'll need to replace the top bound with a variable, and take a limit. Second, we're going to have to antidifferentiate. The integrand is the product of an exponential function, $e^{-x}$, with a polynomial function, $x-1$, so we use integration by parts with $u=x-1$, $\mathrm{d}v=e^{-x}\mathrm{d}u$, $\mathrm{d}u=\mathrm{d}x$, and $v = -e^{-x}$.

$$\begin{aligned} \int \dfrac{x-1}{e^x}~\mathrm{d}x&=-(x-1)e^{-x} + \int e^{-x}~\mathrm{d}x \\&=-(x-1)e^{-x} -e^{-x}+C =-xe^{-x}+C\\ \text{So,}\qquad\int_0^\infty \dfrac{x-1}{e^x}~\mathrm{d}x & = \lim_{b \to \infty}\int_0^b \dfrac{x-1}{e^x}~\mathrm{d}x\\ & = \lim_{b \to \infty}\left[-\frac{x}{e^x}\right]_0^b = \lim_{b \to \infty}\bigg[-\underbrace{\frac{b}{e^b}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}}\bigg]\\ & \stackrel{(*)}{=} \lim_{b \to \infty}-\frac{1}{e^b}=0 \end{aligned}$$

(In the equality marked ($*$), we used l'Hopital's rule.)

So, $\displaystyle\int_0^\infty \dfrac{x-1}{e^x}~\mathrm{d}x=0$.

Remark: this shows that, interestingly, $\displaystyle\int_0^\infty \dfrac{x}{e^x}~\mathrm{d}x=\displaystyle\int_0^\infty \dfrac{1}{e^x}~\mathrm{d}x$.

We see that we have two functions multiplied, but they don't simplify nicely with each other. However, if we differentiate logarithm, and integrate $x^2$, we'll get a polynomial. So, let's use integration by parts.

$$u=\log x \, \mathrm{d}v=x^2\mathrm{d}x$$ $$\mathrm{d}u=(1/x)\mathrm{d}x \, v=x^3/3$$

First, let's antidifferentiate. We'll deal with the limits of integration later.

$$\begin{aligned} \int x^2\log x \mathrm{d}x &=\underbrace{(\log x)}_{u}\underbrace{(x^3/3)}_{v})-\int\underbrace{(x^3/3)}_{v}\underbrace{(1/x)\mathrm{d}x}_{\mathrm{d}u}\\ &=\frac{1}{3}x^3\log x - \frac{1}{3}\int x^2\mathrm{d}x\\ &=\frac{1}{3}x^3\log x -\frac{1}{3}\cdot\frac{1}{3}x^3+C \\ &=\frac{1}{3}x^3\log x - \frac{1}{9}x^3+C \\ \text{ We use the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral.} \\ \int_1^2 x^3\log x \mathrm{d}x&= \left[\frac{1}{3}x^3\log x - \frac{1}{9}x^3\right]_1^2\\ &=\left[\frac{1}{3}2^3\log 2 - \frac{1}{9}2^3 \right] - \left[\frac{1}{3}1^3\log 1 - \frac{1}{9}1^3 \right]\\ &=\frac{8\log 2}{3}-\frac{8}{9}-0+\frac{1}{9}\\ &=\frac{8}{3}\log2-\frac{7}{9} \end{aligned}$$

The derivative of the denominator shows up in the numerator, only differing by a constant, so we perform a substitution. Specifically, substitute $u=x^2-3$, $\mathrm{d}u=2x\,\mathrm{d}x$. This gives

$$\begin{aligned} \int\frac{x}{x^2-3}\mathrm{d}x =\int\frac{\mathrm{d}u/2}{u} =\frac{1}{2}\log|u| + C =\frac{1}{2}\log\big|x^2-3\big| + C \end{aligned}$$

(a) Although a quadratic under a square root often suggests trigonometric substitution, in this case we have an easier substitution. Specifically, let $y=9+x^2$. Then $\mathrm{d}y=2x\mathrm{d}x,\ x\mathrm{d}x=\frac{\mathrm{d}y}{2},\ y(0)=9,$ and $y(4)=25$.