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

The integral from 0 to 2 of (2x-6)(4x^2+2) is -48

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

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

- 48

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- 48

= - 48

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What is the integral from 0 to 2 of (2x-6)(4x^2+2) ?
The integral from 0 to 2 of (2x-6)(4x^2+2) is -48