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

The integral from 0 to 1 of (x^2-1)e^x is -1

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

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

- 1

Solution steps

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

Apply Integration By Parts

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

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

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

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

Expand

(x^{2} - 1) e^{x} - 2 (e^{x} x - e^{x}) : e^{x} x^{2} - 2 e^{x} x + e^{x}

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

Compute the boundaries:

- 1

= - 1

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