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

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

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integral from-1 to 1 of e^xcos(x)

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

\frac{e ^{2} sin ( 1 ) + e ^{2} cos ( 1 ) + sin ( 1 ) - cos ( 1 )}{2 e}

Decimal Notation

1.93342 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{e sin ( 1 ) + e cos ( 1 )}{2} - \frac{- sin ( 1 ) + cos ( 1 )}{2 e}

Simplify

= \frac{e ^{2} sin ( 1 ) + e ^{2} cos ( 1 ) + sin ( 1 ) - cos ( 1 )}{2 e}

\frac{d}{d x} (x^{\frac{9}{5}} y^{7} z^{\frac{2}{7}})

y^{^{'}} = \frac{1}{10} y (500 - y)

x^{5}, x = 2

y = x^{2} + 3 x + 11

\frac{3}{1 + 3 x}

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