What is the integral from-1 to 2 of (3u-8)(u+3) ?

The integral from-1 to 2 of (3u-8)(u+3) is -123/2

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integral from-1 to 2 of (3u-8)(u+3)

\int_{- 1}^{2} (3 u - 8) (u + 3) d u

- \frac{123}{2}

Decimal

- 61.5

Solution steps

\int_{- 1}^{2} (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}^{2}

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

\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}^{2}

Compute the boundaries:

- \frac{123}{2}

= - \frac{123}{2}

\int_{0}^{4} 3 x^{2} d x

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\int_{1}^{5} \frac{1}{t} d t

\int_{1}^{b} 3 c x^{2} + 1 d x

What is the integral from-1 to 2 of (3u-8)(u+3) ?
The integral from-1 to 2 of (3u-8)(u+3) is -123/2