Question

The value of the integral ${\int }_0^{41\frac{\pi }{4}}|cosx|dxis$

• A. $20-\frac{1}{\sqrt{2}}$
• B. $20+\frac{1}{\sqrt{2}}$
• C. $19+\frac{1}{\sqrt{2}}$
• D. $19-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $20+\frac{1}{\sqrt{2}}$

$={\int }_0^{41\frac{\pi }{4}}|cosx|dx={\int }_0^{10\pi }|cosx|dx+={\int }_{10\pi }^{41\frac{\pi }{4}}|cosx|dx$
$=10{\int }_0^{\pi }|cosx|dx+{\int }_{10\pi }^{10\pi +\frac{\pi }{4}}|cosx|dx$
[since |cos x| is a periodic cunction of period $\pi$ ]
$=10{\int }_0^{\pi }|cosx|dx+{\int }_0^{\frac{\pi }{4}}|cosx|dx$
$=10\left({\int }_0^{\frac{\pi }{2}}cosxdx-{\int }_{\frac{\pi }{2}}^{\pi }cosxdx\right)+sinx|\begin{array}{c}\frac{\pi }{4}\\ 0\end{array}$
$=10(sinx|\begin{array}{c}\frac{\pi }{2}\\ 0\end{array}-sinx|\begin{array}{c}\pi \\ \frac{\pi }{2}\end{array})+\frac{1}{\sqrt{2}}$
$=10(1+1)+\frac{1}{\sqrt{2}}=20+\frac{1}{\sqrt{2}}$