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

The integral from 0 to ln(2) of 2x^2e^{3x} is 2(14/27+8/3 ln^2(2)-16/9 ln(2))

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

\int_{0}^{l n (2)} 2 x^{2} e^{3 x} d x

2 (\frac{14}{27} + \frac{8}{3} ln^{2} (2) - \frac{16}{9} ln (2))

Decimal

1.13492 \dots

Solution steps

\int_{0}^{l n (2)} 2 x^{2} e^{3 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{l n (2)} x^{2} e^{3 x} d x

Apply Integration By Parts

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

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

\int \frac{2}{3} e^{3 x} x d x = \frac{2}{27} (3 e^{3 x} x - e^{3 x})

= 2 [\frac{1}{3} (e^{3 x} x^{2} - 3 \cdot \frac{2}{27} (3 e^{3 x} x - e^{3 x}))]_{0}^{l n (2)}

3 \cdot \frac{2}{27} (3 e^{3 x} x - e^{3 x}) = \frac{2}{9} (3 e^{3 x} x - e^{3 x})

= 2 [\frac{1}{3} (e^{3 x} x^{2} - \frac{2}{9} (3 e^{3 x} x - e^{3 x}))]_{0}^{l n (2)}

Compute the boundaries:

\frac{14}{27} + \frac{8}{3} ln^{2} (2) - \frac{16}{9} ln (2)

= 2 (\frac{14}{27} + \frac{8}{3} ln^{2} (2) - \frac{16}{9} ln (2))

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What is the integral from 0 to ln(2) of 2x^2e^{3x} ?
The integral from 0 to ln(2) of 2x^2e^{3x} is 2(14/27+8/3 ln^2(2)-16/9 ln(2))