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

The integral from 0 to 1 of (x-x^2)e^{3x} is (e^3+5)/(27)

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

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

\frac{e ^{3} + 5}{27}

Decimal

0.92909 \dots

Solution steps

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

Expand

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

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

\int_{0}^{1} e^{3 x} x d x = \frac{2 e ^{3} + 1}{9}

\int_{0}^{1} e^{3 x} x^{2} d x = \frac{5 e ^{3} - 2}{27}

= \frac{2 e ^{3} + 1}{9} - \frac{5 e ^{3} - 2}{27}

Simplify

\frac{2 e ^{3} + 1}{9} - \frac{5 e ^{3} - 2}{27} : \frac{e ^{3} + 5}{27}

= \frac{e ^{3} + 5}{27}

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What is the integral from 0 to 1 of (x-x^2)e^{3x} ?
The integral from 0 to 1 of (x-x^2)e^{3x} is (e^3+5)/(27)