What is the integral from 0 to pi of (8e^x+7sin(x)) ?

The integral from 0 to pi of (8e^x+7sin(x)) is 8e^pi+6

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integral from 0 to pi of (8e^x+7sin(x))

\int_{0}^{π} (8 e^{x} + 7 sin (x)) d x

8 e^{π} + 6

Solution steps

\int_{0}^{π} 8 e^{x} + 7 sin (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} 8 e^{x} d x + \int_{0}^{π} 7 sin (x) d x

\int_{0}^{π} 8 e^{x} d x = 8 (e^{π} - 1)

\int_{0}^{π} 7 sin (x) d x = 14

= 8 (e^{π} - 1) + 14

Simplify

8 (e^{π} - 1) + 14 : 8 e^{π} + 6

= 8 e^{π} + 6

\frac{5 x ^{2} + 3 x}{2 x}

y = (x^{3} - 25 x)^{14}, (- 5, 0)

n = 0 \sum \infty \frac{1}{( n + 1 )}

n = 1 \sum \infty \frac{n !}{n 7 ^{n}}

What is the integral from 0 to pi of (8e^x+7sin(x)) ?
The integral from 0 to pi of (8e^x+7sin(x)) is 8e^pi+6