Question

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

• A. 1
• B. –1
• C. 2

Answer 1

The correct option is B –1

The given integral is ${\int }_0^{\frac{\pi }{2}}[\cos x-\sin x]e^{x}dx$
$={\int }_0^{\frac{\pi }{2}}[\cos x+(-\sin x\left)\right]e^{x}dx$
Now, if f(x) = cos x, then, f'(x) = - sin x
The above integral can be written as ${\int }_0^{\frac{\pi }{2}}{e}^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$
The value of the integral is ${\left[e^{x}f\right(x\left)\right]}_0^{\frac{\pi }{2}}$
$=[e^{x}cosx{]}_0^{\frac{\pi }{2}}$
$=e^{\frac{\pi }{2}}\times \cos \frac{\pi }{2}-e^0\times \cos 0$
=$0-1=-1$