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

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

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

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

9 e^{π} + 9

Decimal Notation

217.26623 \dots

Solution steps

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

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

\int_{0}^{π} 9 sin (x) d x = 18

= 9 (e^{π} - 1) + 18

Simplify

9 (e^{π} - 1) + 18 : 9 e^{π} + 9

= 9 e^{π} + 9

\int_{\frac{π}{2}}^{π} x^{2} sin (\frac{x}{2}) d x

\int_{0}^{4} (5 + 7 x) d x

\int_{- 5}^{5} (8 - ∣ x ∣) d x

\int_{1}^{4} (2 x^{2}) d x

\int_{0}^{4} \frac{1}{( x - 2 ) ^{2}} d x

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