What is the integral from 0 to 2 of x(3x-c) ?

The integral from 0 to 2 of x(3x-c) is 2(-c+4)

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integral from 0 to 2 of x(3x-c)

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

2 (- c + 4)

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

2 (- c + 4)

= 2 (- c + 4)

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