적분 기법 전체 문제

Before we can take our Simpson's rule approximation of $\displaystyle\int_1^x \dfrac{1}{t}~\mathrm{d}t$, we need to know how many intervals to use. That means we need to bound our error, which means we need to bound $\frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}$.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left\{\frac{1}{t}\right\}&=-\frac{1}{t^2}& \frac{\mathrm{d}^2}{\mathrm{d}{t}^2}\left\{\frac{1}{t}\right\}&=\frac{2}{t^3}\\ \frac{\mathrm{d}^3}{\mathrm{d}{t}^3}\left\{\frac{1}{t}\right\}&=-\frac{6}{t^4}& \frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}&=\frac{24}{t^5} \end{aligned}$$

So, over the interval $[1,3]$, $\displaystyle\left|\frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}\right| \leq 24$.

Now, we can find an appropriate $n$ to ensure our error will be be less than 0.1 for any $x$ in $[1,3]$:

$$\begin{aligned} \frac{L(b-a)^5}{180 n^4}&<0 .1 \\\frac{24(x-1)^5}{180 n^4}&< \frac{1}{10} \\n^4 &> \frac{24\cdot (x-1)^5}{18} \\ \text{Because $x-1 \leq 2$ for every $x$ in $[1,3]$, if $n^4 > \dfrac{24\cdot 2^5}{18}$, then $n^4 > \dfrac{24\cdot (x-1)^5}{18}$ for every allowed $x$.} \\ n^4&> \frac{24\cdot 2^5}{18}= \frac{128}{3}\\ n&>\sqrt[4]{\frac{128}{3}}\approx 2.6 \end{aligned}$$

Since $n$ must be even, $n=4$ is enough intervals to guarantee our error is not too high for any $x$ in $[1,3]$. Now we find our Simpson's rule approximation with $n=4$, $a=1$, $b=x$, and $\Delta x = \dfrac{x-1}{4}$. The points where we evaluate $\frac{1}{t}$ are:

$$\begin{aligned} x_0 =& 1 & x_1&=1+ \frac{x-1}{4}& x_2&=1+2\frac{x-1}{4}& x_3&=1+3\frac{x-1}{4}& x_4&=1+4\frac{x-1}{4}\\ & & &=\frac{x+3}{4}& &=\frac{x+1}{2}&&=\frac{3x+1}{4}& &=x \end{aligned}$$

$$\begin{aligned} \log x=\int_1^x \frac{1}{t}~\mathrm{d}t&\approx \frac{\Delta x}{3}\left[\frac{1}{x_0}+\frac{4}{x_1}+\frac{2}{x_2}+\frac{4}{x_3}+\frac{1}{x_4}\right]\\ &=\frac{x-1}{12}\left[1+\frac{16}{x+3}+\frac{4}{x+1}+\frac{16}{3x+1}+\frac{1}{x}\right]\\ &=f(x) \end{aligned}$$

Below is a graph of our approximation $f(x)$ and natural logarithm on the same axes. The natural logarithm function is shown red and dashed, while our approximating function is solid blue. Our approximation appears to be quite accurate for small, positive values of $x$.

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.)

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 $y(4)=25$.

$$\begin{aligned} \int_0^4\frac{x}{\sqrt{9+x^2}}\,\mathrm{d}x &=\int_9^{25}\frac{1}{\sqrt{y}}\,\frac{\mathrm{d}y}{2} =\frac{1}{2}\cdot\frac{\sqrt{y}}{1/2}\Big|_9^{25} =5-3= 2 \end{aligned}$$

(b) The power of cosine is odd, so we can reserve one cosine for the differential and change the rest to sines. Substituting $y=\sin x,$ $\mathrm{d}y=\cos x$, $\mathrm{d}x,$ $y(0)=0,$ $y(\pi/2)=1,$ $\cos^2x=1-y^2$:

$$\begin{aligned} \int_0^{\pi/2}\cos^3x\ \sin^2x\,\mathrm{d}x &=\int_0^{\pi/2}\cos^2x\ \sin^2x\ \cos x\,\mathrm{d}x =\int_0^1\!(1-y^2)y^2\,\mathrm{d}y =\int_0^1(y^2-y^4)\,\mathrm{d}y \\ &=\bigg[\frac{y^3}{3}-\frac{y^5}{5}\bigg]_0^1 =\frac{1}{3}-\frac{1}{5} \\ &= \frac{2}{15} \end{aligned}$$

(c) The integrand is the product of two different kinds of functions, with no obvious substitution or simplification. If we differentiate $\log x$, it will match better with the polynomial nature of the rest of the integrand. So, integrate by parts with $u(x)=\log x$ and $\mathrm{d}v=x^3\,\mathrm{d}x$, then $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=x^4/4$.

$$\begin{aligned} \int_1^{e}x^3\log x\,\mathrm{d}x &=\frac{x^4}{4}\log x\bigg|_1^e -\int_1^{e}\frac{x^4}{4}\cdot\frac{1}{x}\,\mathrm{d}x =\frac{e^4}{4}-\int_1^{e}\frac{x^3}{4}\,\mathrm{d}x =\frac{e^4}{4}-\frac{x^4}{16}\bigg|_1^e \\ &=\frac{3e^4}{16}+\frac{1}{16} \end{aligned}$$

(a) Integrate by parts with $u=x$ and $\mathrm{d}v=\sin x\,\mathrm{d}x$ so that $\mathrm{d}u=\mathrm{d}x$ and $v=-\cos x$.

$$\begin{aligned} \int x\sin x\,\mathrm{d}x&= -x\cos x-\int (-\cos x)\,\mathrm{d}x =-x\cos x+\sin x+C\\ \text{So, }\quad\int_0^{\pi/2} x\sin x\,\mathrm{d}x&=\Big[-x\cos x+\sin x\Big]_0^{\pi/2} =1 \end{aligned}$$

(b) The power of cosine is odd, so we can reserve one cosine for $\mathrm{d}u$ and change the rest into sines. Make the substitution $u=\sin x$, $\mathrm{d}u=\cos x\,\mathrm{d}x$.

$$\begin{aligned} \int_0^{\pi/2} \cos^5 x\,\mathrm{d}x &=\int_0^{\pi/2} \big(1-\sin^2x\big)^2\cos x\,\mathrm{d}x =\int_0^1\big(1-u^2\big)^2\,\mathrm{d}u =\int_0^1\big(1-2u^2+u^4\big)\,\mathrm{d}u\\ &=\left[u-\frac{2}{3}u^3+\frac{1}{5}u^5\right]_0^1 =1-\frac{2}{3}+\frac{1}{5} =\frac{8}{15} \end{aligned}$$