What is the integral from 0 to 2 of (6x-3)(8x^2+6) ?

The integral from 0 to 2 of (6x-3)(8x^2+6) is 164

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integral from 0 to 2 of (6x-3)(8x^2+6)

\int_{0}^{2} (6 x - 3) (8 x^{2} + 6) d x

164

Solution steps

\int_{0}^{2} (6 x - 3) (8 x^{2} + 6) d x

Apply Integration By Parts

: [16 x^{4} + 36 x^{2} - 8 x^{3} - 18 x - \int 6 (\frac{8 x ^{3}}{3} + 6 x) d x]_{0}^{2}

= [16 x^{4} + 36 x^{2} - 8 x^{3} - 18 x - \int 6 (\frac{8 x ^{3}}{3} + 6 x) d x]_{0}^{2}

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

= [16 x^{4} + 36 x^{2} - 8 x^{3} - 18 x - 6 (\frac{2 x ^{4}}{3} + 3 x^{2})]_{0}^{2}

Expand

16 x^{4} + 36 x^{2} - 8 x^{3} - 18 x - 6 (\frac{2 x ^{4}}{3} + 3 x^{2}) : 12 x^{4} - 8 x^{3} + 18 x^{2} - 18 x

= [12 x^{4} - 8 x^{3} + 18 x^{2} - 18 x]_{0}^{2}

Compute the boundaries:

164

= 164

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What is the integral from 0 to 2 of (6x-3)(8x^2+6) ?
The integral from 0 to 2 of (6x-3)(8x^2+6) is 164