Integral Calculus (Integration) - SS3 Mathematics Past Questions and Answers - page 1

Let $u=x^2\therefore \frac{du}{dx}=2x\therefore dx=\frac{du}{2x}$

Substituting we have,

${\int }_{}^{}\frac{1}{x^2}dx={\int }_{}^{}\frac{1}{u}.\frac{du}{2x}=\frac{1}{2x}(\ln u)+C$, substitute $u$

$$=\frac{1}{2x}\ln x^2+C$$

$\therefore {\int }_{}^{}\frac{1}{x^2}dx=\frac{\ln x^2}{2x}+C$

Let $u=x^2\therefore \frac{du}{dx}=2x\therefore dx=\frac{du}{2x}$

Substituting we have,

${\int }_{}^{}\frac{1}{x^2}dx={\int }_{}^{}\frac{1}{u}.\frac{du}{2x}=\frac{1}{2x}(\ln u)+C$, substitute $u$

$$=\frac{1}{2x}\ln x^2+C$$

$\therefore {\int }_{}^{}\frac{1}{x^2}dx=\frac{\ln x^2}{2x}+C$

$${\int }_{}^{}udv=uv-{\int }_{}^{}vdu$$

$u=(x-3)$, $\frac{du}{dx}=1$

$\frac{dv}{dx}=(x+3)$, $v=\frac{x^2}{2}+3x$

${\int }_{}^{}(x-3)(x+3)dx=(x-3)(\frac{x^2}{2}+3x)-{\int }_{}^{}(\frac{x^2}{2}+3x)(1)$

$$=(x-3)(x^2+6x)-{\int }_{}^{}{x}^2+6x$$

$$=x^3+3x^2-18x-\frac{x^3}{3}+\frac{6x^2}{2}$$

$$=x^3+3x^2-18x-\frac{x^3}{3}+3x^2$$

$$=x^3+6x^2-18x-\frac{x^3}{3}$$

$$=3x^3+18x^2-54x-x^3$$

$$=2x^3+18x^2-54x$$

$=2x[x^2+9x-27]$

${\int }_1^{2}2x^{3}dx={\left[\frac{2x^4}{4}\right]}_1^2={\left[\frac{x^4}{2}\right]}_1^2=\left[\frac{{(2)}^4}{2}\right]-\left[\frac{(1{)}^4}{2}\right]=\frac{16}{2}-\frac{1}{2}=\frac{15}{2}=7.5$

First integrate ${\int }_{}^{}{e}^{\frac{x}{2}}dx$, let $u=\frac{x}{2}$, $\frac{du}{dx}=\frac{2(1)-x(0)}{4}=\frac{2}{4}=\frac{1}{2}$, $dx=4du$

$${\int }_{}^{}{e}^{u}4du=4{\int }_{}^{}{e}^{u}du=4e^u=4e^{\frac{x}{2}}$$

$${\int }_0^2{e}^{\frac{x}{2}}dx={\left[4e^{\frac{x}{2}}\right]}_0^2=\left[4e^{\frac{2}{2}}\right]-\left[4e^{\frac{0}{2}}\right]=4e-4(1)=4e-4=4(e-1)$$

$${\int }_{}^{}udv=uv-{\int }_{}^{}vdu$$

$u=x$, $du=1$

$dv=\sin x$, $v=-\cos x$

$${\int }_{}^{}x\sin xdx=-x\cos x-{\int }_{}^{}-\cos x$$

$$=-x\cos x+{\int }_{}^{}\cos x$$

$$=-x\cos x+\sin x$$

$$=\sin x-x\cos x$$