Question

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

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is C

-1

${\int }_0^{\frac{\pi }{2}}[cosx+(-sinx\left)\right]e^{x}dx$

Using the property $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$

${\int }_0^{\frac{\pi }{2}}[cosx+(-sinx\left)\right]e^{x}dx$

$=(e^{x}cosx{)}_0^{\frac{\pi }{2}}=-1$