What is the integral from 0 to 1 of sin(x)e^x ?

The integral from 0 to 1 of sin(x)e^x is (-ecos(1)+esin(1)+1)/2

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integral from 0 to 1 of sin(x)e^x

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

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

Decimal Notation

0.90933 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\frac{1}{s} - \frac{s + 8}{s ^{2} + 8 s + 25}

\int ∣ x - 3 ∣ d x

f (x) = (\frac{x + 2}{x - 2})^{5}

f (x) = x^{2} 1 - x^{2}

\frac{\partial}{\partial x} (3 x e^{x^{2} + y^{2}})

What is the integral from 0 to 1 of sin(x)e^x ?
The integral from 0 to 1 of sin(x)e^x is (-ecos(1)+esin(1)+1)/2