What is the integral from 0 to 1 of (x^2+2)e^{-2x} ?

The integral from 0 to 1 of (x^2+2)e^{-2x} is 5/4-9/(4e^2)

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integral from 0 to 1 of (x^2+2)e^{-2x}

\int_{0}^{1} (x^{2} + 2) e^{- 2 x} d x

\frac{5}{4} - \frac{9}{4 e ^{2}}

Decimal Notation

0.94549 \dots

Solution steps

\int_{0}^{1} (x^{2} + 2) e^{- 2 x} d x

Expand

(x^{2} + 2) e^{- 2 x} : e^{- 2 x} x^{2} + 2 e^{- 2 x}

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

\int_{0}^{1} e^{- 2 x} x^{2} d x = - \frac{5}{4 e ^{2}} + \frac{1}{4}

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

= - \frac{5}{4 e ^{2}} + \frac{1}{4} + 1 - \frac{1}{e ^{2}}

Simplify

- \frac{5}{4 e ^{2}} + \frac{1}{4} + 1 - \frac{1}{e ^{2}} : \frac{5}{4} - \frac{9}{4 e ^{2}}

= \frac{5}{4} - \frac{9}{4 e ^{2}}

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What is the integral from 0 to 1 of (x^2+2)e^{-2x} ?
The integral from 0 to 1 of (x^2+2)e^{-2x} is 5/4-9/(4e^2)