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

The integral from 0 to 2 of (6x-6)(4x^2+8) is 32

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

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

32

Solution steps

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

Apply Integration By Parts

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

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

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

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

Expand

(4 x^{2} + 8) (3 x^{2} - 6 x) - 8 (\frac{3 x ^{4}}{4} - 2 x^{3}) : 6 x^{4} - 8 x^{3} + 24 x^{2} - 48 x

= [6 x^{4} - 8 x^{3} + 24 x^{2} - 48 x]_{0}^{2}

Compute the boundaries:

32

= 32

\int_{3}^{9} 1 - \frac{9}{x} d x

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

\int_{0}^{2} \frac{1}{x ^{2} - 2 x + 1} d x

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

\int_{0}^{3} 2 x (x - 5) d x

What is the integral from 0 to 2 of (6x-6)(4x^2+8) ?
The integral from 0 to 2 of (6x-6)(4x^2+8) is 32