What is the integral from-1 to 3 of (3u-8)(u+3) ?

The integral from-1 to 3 of (3u-8)(u+3) is -64

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integral from-1 to 3 of (3u-8)(u+3)

\int_{- 1}^{3} (3 u - 8) (u + 3) d u

- 64

Solution steps

\int_{- 1}^{3} (3 u - 8) (u + 3) d u

Apply Integration By Parts

: [(3 u - 8) (\frac{u ^{2}}{2} + 3 u) - \int 3 (\frac{u ^{2}}{2} + 3 u) d u]_{- 1}^{3}

= [(3 u - 8) (\frac{u ^{2}}{2} + 3 u) - \int 3 (\frac{u ^{2}}{2} + 3 u) d u]_{- 1}^{3}

\int 3 (\frac{u ^{2}}{2} + 3 u) d u = 3 (\frac{u ^{3}}{6} + \frac{3 u ^{2}}{2})

= [(3 u - 8) (\frac{u ^{2}}{2} + 3 u) - 3 (\frac{u ^{3}}{6} + \frac{3 u ^{2}}{2})]_{- 1}^{3}

Compute the boundaries:

- 64

= - 64

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What is the integral from-1 to 3 of (3u-8)(u+3) ?
The integral from-1 to 3 of (3u-8)(u+3) is -64