What is the integral from 1 to 3 of (x^2+e^{-x}) ?

The integral from 1 to 3 of (x^2+e^{-x}) is 26/3+1/e-1/(e^3)

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integral from 1 to 3 of (x^2+e^{-x})

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

\frac{26}{3} + \frac{1}{e} - \frac{1}{e ^{3}}

Decimal Notation

8.98475 \dots

Solution steps

\int_{1}^{3} x^{2} + 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_{1}^{3} x^{2} d x + \int_{1}^{3} e^{- x} d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{3}}

= \frac{26}{3} + \frac{1}{e} - \frac{1}{e ^{3}}

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What is the integral from 1 to 3 of (x^2+e^{-x}) ?
The integral from 1 to 3 of (x^2+e^{-x}) is 26/3+1/e-1/(e^3)