Question

If $[.]$ denotes the greatest integer function, then

Answer 1

(A) ${\int }_0^{\infty }\left[\frac{2}{e^x}\right]dx={\int }_0^{\ln 2}\left[\frac{2}{e^x}\right]dx+{\int }_{\ln 2}^{\infty }\left[\frac{2}{e^x}\right]dx={\int }_0^{\ln 2}1dx+0=\ln 2$

(B) ${\int }_0^{1.5}\left[x^2\right]dx={\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5}\left[x^2\right]dx$

$[\because \left[x^2\right]=\{\begin{array}{c}\begin{array}{cc}0,& 0\le x<1\end{array}\\ \begin{array}{cc}1,& 1\le x<\sqrt{2}\end{array}\\ \begin{array}{cc}2,& \sqrt{2}\le x<1.5\end{array}\end{array}]$

$={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1.dx+{\int }_{\sqrt{2}}^{1.5}2dx={\left[\left(x\right)\right]}_1^{\sqrt{2}}+{\left[2\left(x\right)\right]}_{\sqrt{2}}^{3/2}$

$=\sqrt{2}-1+2(\frac{3}{2}-\sqrt{2})=2-\sqrt{2}.$

(C) Let $P=\frac{\sin x}{2+\cos x}$

$\Rightarrow \frac{dP}{dx}=\frac{(2+\cos x).\cos x-\sin x.(-\sin x)}{{(2+\cos x)}^2}=\frac{2\cos x+1}{{(2+\cos x)}^2}.$

Integrating both sides with respect to $x$ between the limits $0$ and $\frac{\pi }{2},$

we get ${\left[P\right]}_0^{\pi /2}={\int }_0^{\pi /2}\frac{2\cos x+1}{{(2+\cos x)}^2}dx$

i.e., ${\int }_0^{\pi /2}\frac{2\cos x+1}{{(2+\cos x)}^2}dx={\left[\frac{\sin x}{2+\cos x}\right]}_0^{\pi /2}=(\frac{1}{2}-0)=\frac{1}{2}.$

(D) Let $I={\int }_3^4\frac{\left[x^2\right]}{[x^2-14x+49]+\left[x^2\right]}dx$ ...(1)

$={\int }_3^4\frac{\left[x^2\right]}{\left[{(7-x)}^2\right]+\left[x^2\right]}dx={\int }_3^4\frac{\left[{(7-x)}^2\right]}{\left[{(7-(7-x))}^2\right]+\left[{(7-x)}^2\right]}dx$

$[$Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx]$

$={\int }_3^4\frac{\left[{(7-x)}^2\right]}{\left[x^2\right]+\left[{(7-x)}^2\right]}dx$ ...(2)

Adding (1) and (2), we get

$2I={\int }_3^{4}1dx=1.$

$\therefore I=\frac{1}{2}.$