What is the integral from 0 to 2 of (6x-6)(4x^2+3) ?

The integral from 0 to 2 of (6x-6)(4x^2+3) is 32

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integral from 0 to 2 of (6x-6)(4x^2+3)

\int_{0}^{2} (6 x - 6) (4 x^{2} + 3) d x

32

Solution steps

\int_{0}^{2} (6 x - 6) (4 x^{2} + 3) d x

Apply Integration By Parts

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

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

Expand

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

= [6 x^{4} - 8 x^{3} + 9 x^{2} - 18 x]_{0}^{2}

Compute the boundaries:

32

= 32

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

\int_{- 5}^{5} (2 x + 10) d x

\int_{1}^{16} \frac{5}{x} d x

\int_{- \frac{a}{2}}^{\frac{a}{2}} 1 d x

\int_{0}^{4} \frac{1}{x + 1} d x

What is the integral from 0 to 2 of (6x-6)(4x^2+3) ?
The integral from 0 to 2 of (6x-6)(4x^2+3) is 32