What is the integral from 0 to 3 of 8x(2x^2+3) ?

The integral from 0 to 3 of 8x(2x^2+3) is 432

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integral from 0 to 3 of 8x(2x^2+3)

\int_{0}^{3} 8 x (2 x^{2} + 3) d x

432

Solution steps

\int_{0}^{3} 8 x (2 x^{2} + 3) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{0}^{3} x (2 x^{2} + 3) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (2 x^{2} + 3) - \int 2 x^{3} d x]_{0}^{3}

= 8 [\frac{1}{2} x^{2} (2 x^{2} + 3) - \int 2 x^{3} d x]_{0}^{3}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

= 8 [\frac{1}{2} x^{2} (2 x^{2} + 3) - \frac{x ^{4}}{2}]_{0}^{3}

Simplify

8 [\frac{1}{2} x^{2} (2 x^{2} + 3) - \frac{x ^{4}}{2}]_{0}^{3} : 8 [\frac{1}{2} (x^{4} + 3 x^{2})]_{0}^{3}

= 8 [\frac{1}{2} (x^{4} + 3 x^{2})]_{0}^{3}

Compute the boundaries:

54

= 8 \cdot 54

Simplify

= 432

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What is the integral from 0 to 3 of 8x(2x^2+3) ?
The integral from 0 to 3 of 8x(2x^2+3) is 432