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

The integral from 0 to 1 of (x^2+4)e^{-x} is -9/e+6

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integral from 0 to 1 of (x^2+4)e^{-x}

\int_{0}^{1} (x^{2} + 4) e^{- x} d x

- \frac{9}{e} + 6

Decimal Notation

2.68908 \dots

Solution steps

\int_{0}^{1} (x^{2} + 4) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x^{2} + 4) - \int - 2 e^{- x} x d x]_{0}^{1}

= [- e^{- x} (x^{2} + 4) - \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} + 4) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

[- e^{- x} (x^{2} + 4) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 6 e^{- x}]_{0}^{1}

= [- e^{- x} x^{2} - 2 e^{- x} x - 6 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{9}{e} + 6

= - \frac{9}{e} + 6

t \cdot e^{t}

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

cos (f (x) - 2)

\frac{1}{t ^{3}}

11

What is the integral from 0 to 1 of (x^2+4)e^{-x} ?
The integral from 0 to 1 of (x^2+4)e^{-x} is -9/e+6