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

The integral from 0 to 2 of (x^2+3x-1)e^{2x} is 3e^4+1

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

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

3 e^{4} + 1

Decimal Notation

164.79445 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int e^{2 x} x + \frac{3}{2} e^{2 x} d x = \frac{1}{2} e^{2 x} x + \frac{1}{2} e^{2 x}

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

Expand

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

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

Compute the boundaries:

3 e^{4} + 1

= 3 e^{4} + 1

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What is the integral from 0 to 2 of (x^2+3x-1)e^{2x} ?
The integral from 0 to 2 of (x^2+3x-1)e^{2x} is 3e^4+1