What is the integral from 0 to 1 of (x^2+6)e^{-x} ?

The integral from 0 to 1 of (x^2+6)e^{-x} is -(11)/e+8

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integral from 0 to 1 of (x^2+6)e^{-x}

\int_{0}^{1} (x^{2} + 6) e^{- x} d x

- \frac{11}{e} + 8

Solution steps

\int_{0}^{1} (x^{2} + 6) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x^{2} + 6) - \int - 2 e^{- x} x d x]_{0}^{1}

= [- e^{- x} (x^{2} + 6) - \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} + 6) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

[- e^{- x} (x^{2} + 6) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 8 e^{- x}]_{0}^{1}

= [- e^{- x} x^{2} - 2 e^{- x} x - 8 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{11}{e} + 8

= - \frac{11}{e} + 8

y = \frac{1}{x - 5}

\int_{0}^{l n (4)} e^{- 2 x} d x

\frac{d}{d x} (e^{t a n (x)})

\frac{d}{d x} (x 400 - x^{2})

5 y^{^{'}} + 15 y = 375, y (0) = 0

What is the integral from 0 to 1 of (x^2+6)e^{-x} ?
The integral from 0 to 1 of (x^2+6)e^{-x} is -(11)/e+8