What is the integral from 0 to pi of 2e^xsin(x) ?

The integral from 0 to pi of 2e^xsin(x) is e^pi+1

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integral from 0 to pi of 2e^xsin(x)

\int_{0}^{π} 2 e^{x} sin (x) d x

e^{π} + 1

Solution steps

\int_{0}^{π} 2 e^{x} sin (x) d x

Compute the indefinite integral:

\int 2 e^{x} sin (x) d x = - e^{x} cos (x) + e^{x} sin (x) + C

Compute the boundaries:

\int_{0}^{π} 2 e^{x} sin (x) d x = e^{π} - (- 1)

= e^{π} - (- 1)

Simplify

= e^{π} + 1

\int_{e^{3}}^{e^{5}} \frac{1}{x ln ( x )} d x

\int_{0}^{x} 2 x + 1 d x

\int_{\frac{π}{6}}^{\frac{π}{4}} ln (sin (x)) d x

\int_{- 1}^{1} x ∣ 2 x - 1 ∣ d x

\int_{0}^{1} (x + 1) e^{2 x} d x

What is the integral from 0 to pi of 2e^xsin(x) ?
The integral from 0 to pi of 2e^xsin(x) is e^pi+1