What is the integral from-1 to 2 of w^2e^{4w} ?

The integral from-1 to 2 of w^2e^{4w} is e^8-(7e^8)/(32)-(13)/(32e^4)

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integral from-1 to 2 of w^2e^{4w}

\int_{- 1}^{2} w^{2} e^{4 w} d w

e^{8} - \frac{7 e ^{8}}{32} - \frac{13}{32 e ^{4}}

Solution steps

\int_{- 1}^{2} w^{2} e^{4 w} d w

Apply Integration By Parts

: [\frac{1}{4} e^{4 w} w^{2} - \int \frac{1}{2} e^{4 w} w d w]_{- 1}^{2}

= [\frac{1}{4} e^{4 w} w^{2} - \int \frac{1}{2} e^{4 w} w d w]_{- 1}^{2}

\int \frac{1}{2} e^{4 w} w d w = \frac{1}{32} (4 e^{4 w} w - e^{4 w})

= [\frac{1}{4} e^{4 w} w^{2} - \frac{1}{32} (4 e^{4 w} w - e^{4 w})]_{- 1}^{2}

Compute the boundaries:

e^{8} - \frac{7 e ^{8}}{32} - \frac{13}{32 e ^{4}}

= e^{8} - \frac{7 e ^{8}}{32} - \frac{13}{32 e ^{4}}

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What is the integral from-1 to 2 of w^2e^{4w} ?
The integral from-1 to 2 of w^2e^{4w} is e^8-(7e^8)/(32)-(13)/(32e^4)