What is the integral from 0 to 2 of (2x-9)(8x^2+2) ?

The integral from 0 to 2 of (2x-9)(8x^2+2) is -156

Popular >

integral from 0 to 2 of (2x-9)(8x^2+2)

\int_{0}^{2} (2 x - 9) (8 x^{2} + 2) d x

- 156

Solution steps

\int_{0}^{2} (2 x - 9) (8 x^{2} + 2) d x

Apply Integration By Parts

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

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

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

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

Simplify

[(2 x - 9) (\frac{8 x ^{3}}{3} + 2 x) - 2 (\frac{2 x ^{4}}{3} + x^{2})]_{0}^{2} : [2 x (2 x^{3} - 12 x^{2} + x - 9)]_{0}^{2}

= [2 x (2 x^{3} - 12 x^{2} + x - 9)]_{0}^{2}

Compute the boundaries:

- 156

= - 156

\int_{1}^{2} \frac{3 - 6 x}{x ^{3}} d x

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

\int_{0}^{2} ∣ x (x - 1) ∣ d x

\int_{- 1}^{1} x^{2} + x^{4} sin (x) d x

\int_{0}^{\infty} k t e^{- k t} d t

What is the integral from 0 to 2 of (2x-9)(8x^2+2) ?
The integral from 0 to 2 of (2x-9)(8x^2+2) is -156