What is the integral from-1 to 1 of e^xsin(2x) ?

The integral from-1 to 1 of e^xsin(2x) is (-2e^2cos(2)+e^2sin(2)+2cos(2)+sin(2))/(5e)

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integral from-1 to 1 of e^xsin(2x)

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

\frac{- 2 e ^{2} cos ( 2 ) + e ^{2} sin ( 2 ) + 2 cos ( 2 ) + sin ( 2 )}{5 e}

Decimal Notation

0.95249 \dots

Solution steps

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

Compute the indefinite integral:

\int e^{x} sin (2 x) d x = - \frac{2 e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{5} + C

Compute the boundaries:

\int_{- 1}^{1} e^{x} sin (2 x) d x = \frac{- 2 e cos ( 2 ) + e sin ( 2 )}{5} - \frac{- 2 cos ( 2 ) - sin ( 2 )}{5 e}

= \frac{- 2 e cos ( 2 ) + e sin ( 2 )}{5} - \frac{- 2 cos ( 2 ) - sin ( 2 )}{5 e}

Simplify

= \frac{- 2 e ^{2} cos ( 2 ) + e ^{2} sin ( 2 ) + 2 cos ( 2 ) + sin ( 2 )}{5 e}

\int_{0}^{3} (x e^{2 x}) d x

\int_{- 3}^{3} (x^{7} + x^{2} + 2 x + 1) d x

\int_{0}^{3} \frac{1}{x ^{1.7}} d x

\int_{- 2}^{4} cos (x) d x

\int_{0}^{40} 1 - e^{- \frac{x}{22}} d x

What is the integral from-1 to 1 of e^xsin(2x) ?
The integral from-1 to 1 of e^xsin(2x) is (-2e^2cos(2)+e^2sin(2)+2cos(2)+sin(2))/(5e)