What is the integral from 0 to 2 of (x+1)(3x-4) ?

The integral from 0 to 2 of (x+1)(3x-4) is -2

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integral from 0 to 2 of (x+1)(3x-4)

\int_{0}^{2} (x + 1) (3 x - 4) d x

- 2

Solution steps

\int_{0}^{2} (x + 1) (3 x - 4) d x

Apply Integration By Parts

: [(3 x - 4) (\frac{x ^{2}}{2} + x) - \int 3 (\frac{x ^{2}}{2} + x) d x]_{0}^{2}

= [(3 x - 4) (\frac{x ^{2}}{2} + x) - \int 3 (\frac{x ^{2}}{2} + x) d x]_{0}^{2}

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

= [(3 x - 4) (\frac{x ^{2}}{2} + x) - 3 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2})]_{0}^{2}

Compute the boundaries:

- 2

= - 2

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What is the integral from 0 to 2 of (x+1)(3x-4) ?
The integral from 0 to 2 of (x+1)(3x-4) is -2