Question

1)Evaluate the integral:
i)${\int }_2^3{x}^2.dx$

2)Evaluate the integral:
ii)${\int }_4^9\frac{\sqrt{x}}{(30-x^{\frac{3}{2}}{)}^2}.dx$

3)Evaluate the integral:
iii)${\int }_1^2\frac{x}{(x+1)(x+2)}.dx$

4)Evaluate the integral:
iv)${\int }_0^{\frac{\pi }{4}}{\sin }^{3}2t\cos 2t.dt$

Answer 1

1) ${\int }_2^3{x}^2.dx$

First, we will solve using indefinite integral.

$F\left(x\right)=\int x^2.dx=\frac{x^{2+1}}{2+1}=\frac{x^3}{3}$

Now,

${\int }_2^3{x}^2.dx$

$=F\left(3\right)-F\left(2\right)$

$=\frac{3^2}{3}-\frac{2^3}{3}=\frac{27}{3}-\frac{8}{3}$

$=\frac{19}{3}$

2) ${\int }_4^9\frac{\sqrt{x}}{(30-x^{\frac{3}{2}}{)}^2}.dx$

First, we will solve using indefinite integral.

$\int \frac{\sqrt{x}}{(30-x^{\frac{3}{2}}{)}^2}.dx$

Let $30-x^{\frac{3}{2}}=t$ Differentiating $w.r.t.x$ both sides

$\Rightarrow \frac{-3}{2}{x}^{\frac{1}{2}}=\frac{dt}{dx}$

$\Rightarrow \sqrt{x}dx=\frac{-2dt}{3}$

Therefore

$\int \frac{\sqrt{x}.dx}{(30-x^{\frac{3}{2}}{)}^2}$

$=\frac{-2}{3}\int \frac{dt}{t^2}$

$=\frac{-2}{3}{t}^{-2}.dt$

$=\frac{2}{3t}=\frac{2}{3(30-x^{\frac{3}{2}})}$

$[\because t=(30-x^{\frac{3}{2}})]$

Hence

$F\left(x\right)=\frac{2}{3(30-x^{\frac{3}{2}})}$

${\int }_4^9\frac{\sqrt{x}}{{(30-x^{\frac{3}{2}})}^2}.dx$

$=F\left(9\right)-F\left(4\right)$

$=\frac{2}{3}[\frac{1}{30-(9{)}^{\frac{3}{2}}}-\frac{1}{30-(4{)}^{\frac{3}{2}}}]$

$=\frac{2}{3}[\frac{1}{30-27}-\frac{1}{30-8}]$

$=\frac{2}{3}[\frac{1}{3}-\frac{1}{22}]$

$=\frac{2}{3}\left[\frac{22-3}{3\left(22\right)}\right]$

$=\frac{2}{3}\left[\frac{19}{3\left(22\right)}\right]$

$=\frac{19}{99}$

3) ${\int }_1^2\frac{x}{(x+1)(x+2)}.dx$

$F\left(x\right)=\int \frac{x}{(x+1)(x+2)}.dx$

$\Rightarrow F\left(x\right)=\int \frac{2(x+1)-(x+2)}{(x+1)(x+2)}.dx$

$\Rightarrow F\left(x\right)$

$=\int \frac{2(x+1)}{(x+1)(x+2)}.dx$

$-\int \frac{(x+2)}{(x+1)(x+2)}.dx$

$\Rightarrow F\left(x\right)$

$=\int \frac{2}{(x+2)}.dx-\int \frac{1}{(x+1)}.dx$

$\Rightarrow F\left(x\right)=2\log |x+2|-\log |x+1|$

$\Rightarrow F\left(x\right)=\log |x+2{|}^2-\log |x+1|$

$\Rightarrow F\left(x\right)=\log \left|\frac{(x+2{)}^2}{x+1}\right|$

Now,

${\int }_1^2\frac{x}{(x+1)(x+2)}.dx=F\left(2\right)-F\left(1\right)$

​​​​​​​${\int }_1^2\frac{x}{(x+1)(x+2)}.dx$

$=\log \left|\frac{(2+2{)}^2}{2+1}\right|-\log \left|\frac{(1+2{)}^2}{1+1}\right|$

$=\log \left|\frac{16}{3}\right|-\log \left|\frac{9}{2}\right|$

$=\log |\frac{16}{3}\times \frac{2}{9}|$

$=\log \left|\frac{32}{27}\right|$

4) ${\int }_0^{\frac{\pi }{4}}{\sin }^{3}2t\cos 2t.dt$

$F\left(x\right)=\int {\sin }^{3}2t\cos 2t.dt$

Let $\sin 2t=x$

$F\left(x\right)=\int {\sin }^{3}2t\cos 2t.dt$

Let $\sin 2t=x$

Differentiating $w.r.t.t$

$\Rightarrow 2\cos 2t=\frac{dx}{dt}\Rightarrow \cos 2tdt=\frac{dx}{2}$

$\int {\sin }^{3}2t\cos 2t.dt$

$=\frac{1}{2}\int x^3.dx$

$=\frac{1}{2}\times \left(\frac{x^4}{4}\right)$

$=\frac{x^4}{8}=\frac{(\sin 2t{)}^4}{8}$

Hence,

$F\left(t\right)=\frac{(\sin 2t{)}^4}{8}$

${\int }_0^{\frac{\pi }{4}}{\sin }^{3}2t\cos 2t.dt$

$=F\left(\frac{\pi }{4}\right)-F\left(0\right)$

$=\frac{{\left(\sin \frac{2\pi }{4}\right)}^4}{8}-\frac{\left(\sin 2\right(0){)}^4}{8}$

​​​​​​​$=\frac{{\left(\sin \frac{\pi }{2}\right)}^4}{8}-0$

$=\frac{(1{)}^4}{8}-0$

​​​​​​​$=\frac{1}{8}$