What is the integral from 0 to 2 of (6x-3)(8x^2+4) ?

The integral from 0 to 2 of (6x-3)(8x^2+4) is 152

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integral from 0 to 2 of (6x-3)(8x^2+4)

\int_{0}^{2} (6 x - 3) (8 x^{2} + 4) d x

152

Solution steps

\int_{0}^{2} (6 x - 3) (8 x^{2} + 4) d x

Apply Integration By Parts

: [16 x^{4} + 24 x^{2} - 8 x^{3} - 12 x - \int 6 (\frac{8 x ^{3}}{3} + 4 x) d x]_{0}^{2}

= [16 x^{4} + 24 x^{2} - 8 x^{3} - 12 x - \int 6 (\frac{8 x ^{3}}{3} + 4 x) d x]_{0}^{2}

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

= [16 x^{4} + 24 x^{2} - 8 x^{3} - 12 x - 6 (\frac{2 x ^{4}}{3} + 2 x^{2})]_{0}^{2}

Expand

16 x^{4} + 24 x^{2} - 8 x^{3} - 12 x - 6 (\frac{2 x ^{4}}{3} + 2 x^{2}) : 12 x^{4} - 8 x^{3} + 12 x^{2} - 12 x

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

Compute the boundaries:

152

= 152

\int_{3}^{5} x^{2} + x - 5 d x

\int_{0}^{1} 12 x ln (3 x) d x

\int_{0}^{2} 16 - 2 x d x

\int_{- 1}^{1} ln (x + 2) d x

\int_{0}^{\frac{π}{2}} \frac{1}{cos ^{2} ( x )} d x

What is the integral from 0 to 2 of (6x-3)(8x^2+4) ?
The integral from 0 to 2 of (6x-3)(8x^2+4) is 152