What is the integral from 0 to 2 of (6x-3)(4x^2+7) ?

The integral from 0 to 2 of (6x-3)(4x^2+7) is 106

integral from 0 to 2 of (6x-3)(4x^2+7)

\int_{0}^{2} (6 x - 3) (4 x^{2} + 7) d x

106

Solution steps

\int_{0}^{2} (6 x - 3) (4 x^{2} + 7) d x

Apply Integration By Parts

= [(4 x^{2} + 7) (3 x^{2} - 3 x) - \int 8 x (3 x^{2} - 3 x) d x]_{0}^{2}

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

= [(4 x^{2} + 7) (3 x^{2} - 3 x) - 8 (\frac{3}{4} x^{4} - x^{3})]_{0}^{2}

Expand

(4 x^{2} + 7) (3 x^{2} - 3 x) - 8 (\frac{3}{4} x^{4} - x^{3}) : 6 x^{4} - 4 x^{3} + 21 x^{2} - 21 x

= [6 x^{4} - 4 x^{3} + 21 x^{2} - 21 x]_{0}^{2}

Compute the boundaries:

106

= 106

\int_{- 1}^{1} 2 y^{2} - 2 d y

\int_{2}^{7} \frac{14 x}{x ^{2} + 1} d x

\int_{0}^{l n (6)} e^{5 x} d x

\int_{- 5}^{4} x^{2} d x

\int_{0}^{1} (x^{3} - 3 x^{2} + 2 x) d x

What is the integral from 0 to 2 of (6x-3)(4x^2+7) ?
The integral from 0 to 2 of (6x-3)(4x^2+7) is 106