Question

Let $[t]$ denote the greatest integer less than or equal to t Then the value of the integral $\int\limits_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x$ is equal to

• A. $\frac{52(1- e )}{ e }$
• B. $\frac{52}{ e }$
• C. $\frac{52(2+ e )}{ e }$
• D. $\frac{104}{ e }$

Answer 1

The Correct Option is B

$\underset{-3}{\overset{101}{\int }}\left([\sin \pi x]+e^{[\cos 2\pi x]}\right)dx$
$52\underset{0}{\overset{2}{\int }}\left([\sin \pi x]+e^{[\cos 2\pi x]}\right)dt$
$\frac{52}{\pi }\underset{0}{\overset{2\pi }{\int }}\left([\sin t]+e^{[\cos 2t]}\right)dt$
$\frac{52}{\pi }\underset{0}{\overset{2\pi }{\int }}\left([\sin t]dt+\underset{0}{\overset{2\pi }{\int }}{e}^{[\cos 2t]}dt\right)$
$I_1=\underset{0}{\overset{2\pi }{\int }}[\sin t]dt$
Using King
$I_1=\underset{0}{\overset{2\pi }{\int }}[-\sin t]dt$
$2I_1=\underset{0}{\overset{2\pi }{\int }}(-1)dt=-2\pi$
$I_1=-\pi$
$I_2=2\underset{0}{\overset{\pi }{\int }}{e}^{[\cos 2t]}dt$
$=2.2\underset{0}{\overset{\pi /2}{\int }}{e}^{[\cos 2t]}dt$
$=4\left(\underset{0}{\overset{\pi /4}{\int }}{e}^0\cdot dt+\underset{\pi /4}{\overset{\pi /2}{\int }}{e}^{-1}dt\right)$
$4\left(\frac{\pi }{4}+e^{-1}\left(\frac{\pi }{4}\right)\right)=\pi \left(1+e^{-1}\right)$
$I=\frac{52}{\pi }\left(-\pi +\pi +\pi e^{-1}\right)=\frac{52}{e}$