Question

Evaluate the integral
${\int }_0^{\frac{\pi }{2}}(\cos x-\sin x)e^{x}dx$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

${\int }_0^{\frac{\pi }{2}}(\cos x-\sin x)e^{x}dx$
$=\int e^x(\cos x-\sin x)dx$

It is in the form of
$\int e^x\left(f\right(x)+f^1(x\left)\right)dx=e^{x}f\left(x\right)+c$

Thus $\int e^x(\cos x-\sin x)dx=e^x\cos x+c$
${\int }_0^{\frac{\pi }{2}}{e}^x(\cos x-\sin x)dx=(e^x\cos x+c{)}_0^{\pi /2}$
$=(e^{\frac{\pi }{2}}\cdot \cos \frac{\pi }{2}+c)-(e^0\cos 0+c)$
$=-1$

Thus ${\int }_0^{\frac{\pi }{2}}{e}^x(\cos x-\sin x)dx=-1$