What is the integral from 0 to pi of (9e^x+7sin(x)) ?

The integral from 0 to pi of (9e^x+7sin(x)) is 9e^pi+5

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integral from 0 to pi of (9e^x+7sin(x))

\int_{0}^{π} (9 e^{x} + 7 sin (x)) d x

9 e^{π} + 5

Decimal Notation

213.26623 \dots

Solution steps

\int_{0}^{π} 9 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} 9 e^{x} d x + \int_{0}^{π} 7 sin (x) d x

\int_{0}^{π} 9 e^{x} d x = 9 (e^{π} - 1)

\int_{0}^{π} 7 sin (x) d x = 14

= 9 (e^{π} - 1) + 14

Simplify

9 (e^{π} - 1) + 14 : 9 e^{π} + 5

= 9 e^{π} + 5

\int x \cdot a^{x} d x

\int e^{5 t} cos (t) d t

\int \frac{1}{4 - 49 x ^{2}} d x

y = e^{- k x^{2}}

1 + x^{2}

What is the integral from 0 to pi of (9e^x+7sin(x)) ?
The integral from 0 to pi of (9e^x+7sin(x)) is 9e^pi+5