Fifteen definite integrals

Text

Evaluate the following definite integrals, stating for each the method used (immediate, by parts, by substitution, partial fractions): a) $\int_1^2(x^3-2x^2)\,dx$ b) $\int_0^{\pi/4}\cos(2x)\,dx$ c) $\int_1^2 2x e^x\,dx$ d) $\int_0^{\pi/4}\tan x\,dx$ e) $\int_1^{\pi/4}\tan^2 x\,dx$ f) $\int_1^{\pi/4}\sin^2 x\,dx$ g) $\int_1^2\arctan x\,dx$ h) $\int_{-1}^1\dfrac{x^4+x}{x^2-5x+6}\,dx$ i) $\int_4^5\dfrac{2x+4}{3x^2-6x+3}\,dx$ j) $\int_0^1 x e^{-x^2}\,dx$ k) $\int_0^1\dfrac{t}{1+t^2}\,dt$ l) $\int_1^2\ln t\,dt$ m) $\int_0^{\pi/2}e^x\cos x\,dx$ n) $\int_0^1\sqrt{x}\,dx$ o) $\int_0^1 x\sqrt{x}\,dx$

Solution

a) Immediate. $\int_1^2(x^3-2x^2)\,dx=\left[\dfrac{x^4}{4}-\dfrac{2x^3}{3}\right]_1^2=\left(4-\dfrac{16}{3}\right)-\left(\dfrac14-\dfrac23\right)=-\dfrac{11}{12}\approx-0.9167.$

b) Immediate. $\int_0^{\pi/4}\cos(2x)\,dx=\left[\dfrac{\sin(2x)}{2}\right]_0^{\pi/4}=\dfrac{\sin(\pi/2)}{2}=\dfrac12.$

c) By parts. With $u=2x$, $dv=e^x dx$: an antiderivative is $2e^x(x-1)$. Hence $\int_1^2 2xe^x\,dx=\big[2e^x(x-1)\big]_1^2=2e^2-0=2e^2\approx14.778.$

d) Substitution $u=\cos x$: $\int\tan x\,dx=-\ln|\cos x|$. $\big[-\ln|\cos x|\big]_0^{\pi/4}=-\ln\dfrac{\sqrt2}{2}=\dfrac12\ln 2\approx0.3466.$

e) Immediate using $\tan^2 x=\sec^2 x-1$: antiderivative $\tan x-x$. Note $\pi/4<1$, so the limits are “reversed”. $\big[\tan x-x\big]_1^{\pi/4}=\left(1-\dfrac{\pi}{4}\right)-(\tan 1-1)=2-\dfrac{\pi}{4}-\tan 1\approx-0.3428.$

f) Immediate using $\sin^2 x=\dfrac{1-\cos 2x}{2}$: antiderivative $\dfrac{x}{2}-\dfrac{\sin 2x}{4}$. $\big[\tfrac{x}{2}-\tfrac{\sin 2x}{4}\big]_1^{\pi/4}\approx0.1427-0.2727=-0.1300.$

g) By parts with $u=\arctan x$, $dv=dx$: $\int\arctan x\,dx=x\arctan x-\dfrac12\ln(1+x^2)$. $\big[x\arctan x-\tfrac12\ln(1+x^2)\big]_1^2=\left(2\arctan 2-\tfrac12\ln 5\right)-\left(\tfrac{\pi}{4}-\tfrac12\ln 2\right)\approx0.9708.$

h) Long division + partial fractions. $\dfrac{x^4+x}{x^2-5x+6}=x^2+5x+19-\dfrac{18}{x-2}+\dfrac{84}{x-3}$. Integrating over $[-1,1]$: $\left[\tfrac{x^3}{3}+\tfrac{5x^2}{2}+19x-18\ln|x-2|+84\ln|x-3|\right]_{-1}^1=\tfrac23+38+18\ln 3-84\ln 2\approx0.2156.$

i) Perfect-square denominator. $3x^2-6x+3=3(x-1)^2$, so $\dfrac{2x+4}{3(x-1)^2}=\dfrac{1}{3}\!\left(\dfrac{2}{x-1}+\dfrac{6}{(x-1)^2}\right)$. Integrating over $[4,5]$: $\tfrac13\!\left(2\ln\tfrac43+\tfrac12\right)\approx0.3585.$

j) Substitution $u=x^2$, $du=2x\,dx$: $\int_0^1 xe^{-x^2}dx=\tfrac12\big[-e^{-u}\big]_0^1=\tfrac12(1-e^{-1})\approx0.3161.$

k) Substitution $u=1+t^2$: $\int_0^1\dfrac{t}{1+t^2}dt=\tfrac12\big[\ln(1+t^2)\big]_0^1=\tfrac12\ln 2\approx0.3466.$

l) By parts ($u=\ln t$, $dv=dt$): $\int\ln t\,dt=t\ln t-t$. $\big[t\ln t-t\big]_1^2=(2\ln 2-2)-(-1)=2\ln 2-1\approx0.3863.$

m) By parts (twice): $\int e^x\cos x\,dx=\dfrac{e^x(\sin x+\cos x)}{2}$. $\left[\dfrac{e^x(\sin x+\cos x)}{2}\right]_0^{\pi/2}=\dfrac{e^{\pi/2}\cdot 1}{2}-\dfrac{1}{2}=\dfrac{e^{\pi/2}-1}{2}\approx1.9052.$

n) Immediate. $\int_0^1\sqrt{x}\,dx=\left[\tfrac23 x^{3/2}\right]_0^1=\dfrac23\approx0.6667.$

o) Immediate. $x\sqrt{x}=x^{3/2}$, so $\int_0^1 x^{3/2}dx=\left[\tfrac25 x^{5/2}\right]_0^1=\dfrac25=0.4.$

$\boxed{a=-\tfrac{11}{12},\ b=\tfrac12,\ c=2e^2,\ d=\tfrac12\ln2,\ \dots,\ n=\tfrac23,\ o=\tfrac25}$