Question

(i) $\underset{1}{\overset{4}{\int }}f\left(x\right)dx,\mathrm{where}f\left(x\right)=\left\{\begin{array}{l}4x+3,\mathrm{if}1\le x\le 2\\ 3x+5,\mathrm{if}2\le x\le 4\end{array}\right.$

(ii) $\underset{0}{\overset{9}{\int }}f\left(x\right)dx,\mathrm{where}f\left(x\right)\left\{\begin{array}{ccc}\sin x& ,& 0\le x\le \mathrm{\pi }/2\\ 1& ,& \mathrm{\pi }/2\le \mathrm{x}\le 3\\ e^{x-3}& ,& 3\le x\le 9\end{array}\right.$

(iii) $\underset{1}{\overset{4}{\int }}f\left(x\right)dx,\mathrm{where}f\left(x\right)=\left\{\begin{array}{ccc}7x+3& ,& \mathrm{if}1\le x\le 3\\ 8x& ,& \mathrm{if}3\le x\le 4\end{array}\right.$

(iv) $\underset{-1}{\overset{2}{\int }}\frac{\left|x\right|}{x}dx$

Answer 1

(iv) $\underset{-1}{\overset{2}{\int }}\frac{\left|x\right|}{x}dx$


$$\begin{gathered} \mathrm{Let}I=\underset{-1}{\overset{2}{\int }}\frac{\left|x\right|}{x}dx \\ \mathrm{We}\mathrm{know}, \\ \left|x\right|=\left\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\right. \\ \mathrm{Thus}, \\ I=\underset{-1}{\overset{0}{\int }}\frac{\left|x\right|}{x}dx+\underset{0}{\overset{2}{\int }}\frac{\left|x\right|}{x}dx \\ =\underset{-1}{\overset{0}{\int }}\frac{-x}{x}dx+\underset{0}{\overset{2}{\int }}\frac{x}{x}dx \\ =\underset{-1}{\overset{0}{\int }}-1dx+\underset{0}{\overset{2}{\int }}1dx \\ ={\left[-x\right]}_{-1}^0+{\left[x\right]}_0^2 \\ =\left(-0-1\right)+\left(2-0\right) \\ =-1+2 \\ =1 \end{gathered}$$


Hence, $\underset{-1}{\overset{2}{\int }}\frac{\left|x\right|}{x}dx$ = 1.