evaluate the integral.\n int_{0}^{pi}(7e^{x}+5sin(x))dx

Explanation:

Step1: Use integral sum - rule

$\int_{0}^{\pi}(7e^{x}+5\sin(x))dx=\int_{0}^{\pi}7e^{x}dx+\int_{0}^{\pi}5\sin(x)dx$

Step2: Take out constants

$7\int_{0}^{\pi}e^{x}dx + 5\int_{0}^{\pi}\sin(x)dx$

Step3: Integrate $e^{x}$ and $\sin(x)$

We know that $\int e^{x}dx=e^{x}+C$ and $\int\sin(x)dx=-\cos(x)+C$. So, $7\left[e^{x}\right]{0}^{\pi}+5\left[-\cos(x)\right]{0}^{\pi}$

Step4: Evaluate the definite - integrals

For $7\left[e^{x}\right]{0}^{\pi}$, we have $7(e^{\pi}-e^{0}) = 7(e^{\pi}-1)$. For $5\left[-\cos(x)\right]{0}^{\pi}$, we have $5(-\cos(\pi)+\cos(0))=5(1 + 1)=10$.

Step5: Combine the results

$7(e^{\pi}-1)+10=7e^{\pi}-7 + 10=7e^{\pi}+3$

Answer:

$7e^{\pi}+3$