What is the integral from 0 to 1 of (x^2+3x-1)e^x ?

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

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integral from 0 to 1 of (x^2+3x-1)e^x

\int_{0}^{1} (x^{2} + 3 x - 1) e^{x} d x

2

Solution steps

\int_{0}^{1} (x^{2} + 3 x - 1) e^{x} d x

Apply Integration By Parts

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

= [(x^{2} + 3 x - 1) e^{x} - \int (2 x + 3) e^{x} d x]_{0}^{1}

\int (2 x + 3) e^{x} d x = 2 e^{x} x + e^{x}

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

Expand

(x^{2} + 3 x - 1) e^{x} - (2 e^{x} x + e^{x}) : e^{x} x^{2} + e^{x} x - 2 e^{x}

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

Compute the boundaries:

2

= 2

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