What is the integral from 0 to 2 of (6x-3)(4x^2+5) ?

The integral from 0 to 2 of (6x-3)(4x^2+5) is 94

Popular >

integral from 0 to 2 of (6x-3)(4x^2+5)

\int_{0}^{2} (6 x - 3) (4 x^{2} + 5) d x

94

Solution steps

\int_{0}^{2} (6 x - 3) (4 x^{2} + 5) d x

Apply Integration By Parts

: [(4 x^{2} + 5) (3 x^{2} - 3 x) - \int 8 x (3 x^{2} - 3 x) d x]_{0}^{2}

= [(4 x^{2} + 5) (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} + 5) (3 x^{2} - 3 x) - 8 (\frac{3}{4} x^{4} - x^{3})]_{0}^{2}

Expand

(4 x^{2} + 5) (3 x^{2} - 3 x) - 8 (\frac{3}{4} x^{4} - x^{3}) : 6 x^{4} - 4 x^{3} + 15 x^{2} - 15 x

= [6 x^{4} - 4 x^{3} + 15 x^{2} - 15 x]_{0}^{2}

Compute the boundaries:

94

= 94

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) - 3 d x

\int_{0}^{\frac{1}{2}} 3 x^{2} d x

\int_{0}^{2} x^{2} 1 + 4 x^{2} d x

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

\int_{1}^{3} 6 x^{2} - 4 x + 5 d x

What is the integral from 0 to 2 of (6x-3)(4x^2+5) ?
The integral from 0 to 2 of (6x-3)(4x^2+5) is 94