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

The integral from 0 to 1 of (x^2+9)e^{-x} is -(14)/e+11

Popular >

integral from 0 to 1 of (x^2+9)e^{-x}

\int_{0}^{1} (x^{2} + 9) e^{- x} d x

- \frac{14}{e} + 11

Decimal

5.84968 \dots

Solution steps

\int_{0}^{1} (x^{2} + 9) e^{- x} d x

Apply Integration By Parts

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

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

Simplify

[- e^{- x} (x^{2} + 9) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 11 e^{- x}]_{0}^{1}

= [- e^{- x} x^{2} - 2 e^{- x} x - 11 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{14}{e} + 11

= - \frac{14}{e} + 11

\int_{0}^{\frac{2}{3}} 4 - 9 x^{2} d x

\int_{0}^{\infty} 12 e^{- 4 x} d x

5 y ln (x) - x y^{^{'}} = 0

e^{2 t}

\frac{\partial}{\partial y} (- e^{- x} z cos (y))

What is the integral from 0 to 1 of (x^2+9)e^{-x} ?
The integral from 0 to 1 of (x^2+9)e^{-x} is -(14)/e+11