What is the integral from 0 to 2 of (6x-6)(8x^2+7) ?

The integral from 0 to 2 of (6x-6)(8x^2+7) is 64

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integral from 0 to 2 of (6x-6)(8x^2+7)

\int_{0}^{2} (6 x - 6) (8 x^{2} + 7) d x

64

Solution steps

\int_{0}^{2} (6 x - 6) (8 x^{2} + 7) d x

Apply Integration By Parts

: [(6 x - 6) (\frac{8 x ^{3}}{3} + 7 x) - \int 6 (\frac{8 x ^{3}}{3} + 7 x) d x]_{0}^{2}

= [(6 x - 6) (\frac{8 x ^{3}}{3} + 7 x) - \int 6 (\frac{8 x ^{3}}{3} + 7 x) d x]_{0}^{2}

\int 6 (\frac{8 x ^{3}}{3} + 7 x) d x = 6 (\frac{2 x ^{4}}{3} + \frac{7 x ^{2}}{2})

= [(6 x - 6) (\frac{8 x ^{3}}{3} + 7 x) - 6 (\frac{2 x ^{4}}{3} + \frac{7 x ^{2}}{2})]_{0}^{2}

Expand

(6 x - 6) (\frac{8 x ^{3}}{3} + 7 x) - 6 (\frac{2 x ^{4}}{3} + \frac{7 x ^{2}}{2}) : 12 x^{4} - 16 x^{3} + 21 x^{2} - 42 x

= [12 x^{4} - 16 x^{3} + 21 x^{2} - 42 x]_{0}^{2}

Compute the boundaries:

64

= 64

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What is the integral from 0 to 2 of (6x-6)(8x^2+7) ?
The integral from 0 to 2 of (6x-6)(8x^2+7) is 64