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

The integral from 0 to 1 of-2(x^2+1)e^{-x} is -2(-6/e+3)

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

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

- 2 (- \frac{6}{e} + 3)

Decimal Notation

- 1.58544 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= - 2 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= - 2 (- \frac{6}{e} + 3)

f (x) = arcsec (2 x)

5 t - 3

\frac{\partial}{\partial x} ((x + y)^{8})

r^{2} - 6 x^{2}

f (x) = x^{4} - 13 x^{2} + 36, at x = 2

What is the integral from 0 to 1 of-2(x^2+1)e^{-x} ?
The integral from 0 to 1 of-2(x^2+1)e^{-x} is -2(-6/e+3)