What is the integral from 0 to 1 of (x^2+7)e^{-x} ?

The integral from 0 to 1 of (x^2+7)e^{-x} is -(12)/e+9

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integral from 0 to 1 of (x^2+7)e^{-x}

\int_{0}^{1} (x^{2} + 7) e^{- x} d x

- \frac{12}{e} + 9

Decimal Notation

4.58544 \dots

Solution steps

\int_{0}^{1} (x^{2} + 7) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x^{2} + 7) - \int - 2 e^{- x} x d x]_{0}^{1}

= [- e^{- x} (x^{2} + 7) - \int - 2 e^{- x} x d x]_{0}^{1}

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

= [- e^{- x} (x^{2} + 7) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

[- e^{- x} (x^{2} + 7) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 9 e^{- x}]_{0}^{1}

= [- e^{- x} x^{2} - 2 e^{- x} x - 9 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{12}{e} + 9

= - \frac{12}{e} + 9

16 y^{^{''}} + 16 y^{^{'}} - 5 y = 0

f (x) = 2 x

\int \frac{x + 8}{x ^{2} + 25} d x

\frac{d}{d x} (x^{2} e^{x^{2}})

\frac{d}{d x} (x^{2} ln (4 x))

What is the integral from 0 to 1 of (x^2+7)e^{-x} ?
The integral from 0 to 1 of (x^2+7)e^{-x} is -(12)/e+9