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

The integral from 0 to 2 of (6x-6)(8x^2+4) is 64

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

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

64

Solution steps

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

Apply Integration By Parts

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

= [(6 x - 6) (\frac{8 x ^{3}}{3} + 4 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})

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

Expand

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

= [12 x^{4} - 16 x^{3} + 12 x^{2} - 24 x]_{0}^{2}

Compute the boundaries:

64

= 64

\int_{- \infty}^{e} 24 e^{- x} d x

\int_{0}^{3} ∣ 5 t - 8 ∣ d t

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

\int_{1}^{4} 4 x d x

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

What is the integral from 0 to 2 of (6x-6)(8x^2+4) ?
The integral from 0 to 2 of (6x-6)(8x^2+4) is 64