What is the integral from 0 to 1 of (x^2+3)e^{-x} ?

The integral from 0 to 1 of (x^2+3)e^{-x} is -8/e+5

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integral from 0 to 1 of (x^2+3)e^{-x}

\int_{0}^{1} (x^{2} + 3) e^{- x} d x

- \frac{8}{e} + 5

Decimal Notation

2.05696 \dots

Solution steps

\int_{0}^{1} (x^{2} + 3) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x^{2} + 3) - \int - 2 e^{- x} x d x]_{0}^{1}

= [- e^{- x} (x^{2} + 3) - \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} + 3) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

[- e^{- x} (x^{2} + 3) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 5 e^{- x}]_{0}^{1}

= [- e^{- x} x^{2} - 2 e^{- x} x - 5 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{8}{e} + 5

= - \frac{8}{e} + 5

\int - y e^{y} d y

\frac{\partial}{\partial x} (x^{2} y + x y)

\int_{0}^{\frac{π}{2}} sin^{4} (x) d x

\int \frac{( ln ( x ) - 1 )}{x ^{2}} d x

\frac{d}{d x} (9 x^{5})

What is the integral from 0 to 1 of (x^2+3)e^{-x} ?
The integral from 0 to 1 of (x^2+3)e^{-x} is -8/e+5