What is the integral from 0 to 2 of (6x-9)(4x^2+6) ?

The integral from 0 to 2 of (6x-9)(4x^2+6) is -36

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integral from 0 to 2 of (6x-9)(4x^2+6)

\int_{0}^{2} (6 x - 9) (4 x^{2} + 6) d x

- 36

Solution steps

\int_{0}^{2} (6 x - 9) (4 x^{2} + 6) d x

Apply Integration By Parts

: [(4 x^{2} + 6) (3 x^{2} - 9 x) - \int 8 x (3 x^{2} - 9 x) d x]_{0}^{2}

= [(4 x^{2} + 6) (3 x^{2} - 9 x) - \int 8 x (3 x^{2} - 9 x) d x]_{0}^{2}

\int 8 x (3 x^{2} - 9 x) d x = 8 (\frac{3 x ^{4}}{4} - 3 x^{3})

= [(4 x^{2} + 6) (3 x^{2} - 9 x) - 8 (\frac{3 x ^{4}}{4} - 3 x^{3})]_{0}^{2}

Expand

(4 x^{2} + 6) (3 x^{2} - 9 x) - 8 (\frac{3 x ^{4}}{4} - 3 x^{3}) : 6 x^{4} - 12 x^{3} + 18 x^{2} - 54 x

= [6 x^{4} - 12 x^{3} + 18 x^{2} - 54 x]_{0}^{2}

Compute the boundaries:

- 36

= - 36

\int_{2}^{4} x arcsec (x) d x

\int_{\frac{π}{6}}^{\frac{π}{3}} sec^{2} (x) d x

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

\int_{0}^{5} x^{4} d x

\int_{0}^{\frac{π}{4}} t^{2} sin (4 t) d t

What is the integral from 0 to 2 of (6x-9)(4x^2+6) ?
The integral from 0 to 2 of (6x-9)(4x^2+6) is -36