What is the integral from 0 to 1 of x^3+xe^{-x} ?

The integral from 0 to 1 of x^3+xe^{-x} is 5/4-2/e

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integral from 0 to 1 of x^3+xe^{-x}

\int_{0}^{1} x^{3} + x e^{- x} d x

\frac{5}{4} - \frac{2}{e}

Decimal Notation

0.51424 \dots

Solution steps

\int_{0}^{1} x^{3} + x e^{- 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}^{1} x^{3} d x + \int_{0}^{1} x e^{- x} d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} x e^{- x} d x = - \frac{2}{e} + 1

= \frac{1}{4} - \frac{2}{e} + 1

Combine the fractions

1 + \frac{1}{4} : \frac{5}{4}

= \frac{5}{4} - \frac{2}{e}

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What is the integral from 0 to 1 of x^3+xe^{-x} ?
The integral from 0 to 1 of x^3+xe^{-x} is 5/4-2/e