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

The integral from 2 to 4 of e^{-x}(x^2+5) is -(31)/(e^4)+(15)/(e^2)

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integral from 2 to 4 of e^{-x}(x^2+5)

\int_{2}^{4} e^{- x} (x^{2} + 5) d x

- \frac{31}{e ^{4}} + \frac{15}{e ^{2}}

Decimal

1.46224 \dots

Solution steps

\int_{2}^{4} e^{- x} (x^{2} + 5) d x

Apply Integration By Parts

: [- e^{- x} (x^{2} + 5) - \int - 2 e^{- x} x d x]_{2}^{4}

= [- e^{- x} (x^{2} + 5) - \int - 2 e^{- x} x d x]_{2}^{4}

\int - 2 e^{- x} x d x = - 2 (- e^{- x} x - e^{- x})

= [- e^{- x} (x^{2} + 5) - (- 2 (- e^{- x} x - e^{- x}))]_{2}^{4}

Simplify

[- e^{- x} (x^{2} + 5) - (- 2 (- e^{- x} x - e^{- x}))]_{2}^{4} : [- e^{- x} x^{2} - 2 e^{- x} x - 7 e^{- x}]_{2}^{4}

= [- e^{- x} x^{2} - 2 e^{- x} x - 7 e^{- x}]_{2}^{4}

Compute the boundaries:

- \frac{31}{e ^{4}} + \frac{15}{e ^{2}}

= - \frac{31}{e ^{4}} + \frac{15}{e ^{2}}

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What is the integral from 2 to 4 of e^{-x}(x^2+5) ?
The integral from 2 to 4 of e^{-x}(x^2+5) is -(31)/(e^4)+(15)/(e^2)