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)

\int 4 cos (8 x) d x

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = e^{2 x}

\int x \cdot 5^{x} d x

\frac{e ^{- s}}{s - 1}

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)