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

The integral from 0 to 3 of (x^2-x+1)e^{2x} is (10e^6-4)/4

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

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

\frac{10 e ^{6} - 4}{4}

Decimal Notation

1007.57198 \dots

Solution steps

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

Expand

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

= \int_{0}^{3} e^{2 x} x^{2} - e^{2 x} x + e^{2 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{3} e^{2 x} x^{2} d x - \int_{0}^{3} e^{2 x} x d x + \int_{0}^{3} e^{2 x} d x

\int_{0}^{3} e^{2 x} x^{2} d x = \frac{13 e ^{6} - 1}{4}

\int_{0}^{3} e^{2 x} x d x = \frac{5 e ^{6} + 1}{4}

\int_{0}^{3} e^{2 x} d x = \frac{e ^{6} - 1}{2}

= \frac{13 e ^{6} - 1}{4} - \frac{5 e ^{6} + 1}{4} + \frac{e ^{6} - 1}{2}

Simplify

\frac{13 e ^{6} - 1}{4} - \frac{5 e ^{6} + 1}{4} + \frac{e ^{6} - 1}{2} : \frac{10 e ^{6} - 4}{4}

= \frac{10 e ^{6} - 4}{4}

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What is the integral from 0 to 3 of (x^2-x+1)e^{2x} ?
The integral from 0 to 3 of (x^2-x+1)e^{2x} is (10e^6-4)/4