What is the integral from 0 to 3 of 7x(x-3) ?

The integral from 0 to 3 of 7x(x-3) is -63/2

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integral from 0 to 3 of 7x(x-3)

\int_{0}^{3} 7 x (x - 3) d x

- \frac{63}{2}

Solution steps

\int_{0}^{3} 7 x (x - 3) d x

Take the constant out:

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

= 7 \cdot \int_{0}^{3} x (x - 3) d x

Expand

x (x - 3) : x^{2} - 3 x

= 7 \cdot \int_{0}^{3} x^{2} - 3 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (\int_{0}^{3} x^{2} d x - \int_{0}^{3} 3 x d x)

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 3 x d x = \frac{27}{2}

= 7 (9 - \frac{27}{2})

Simplify

7 (9 - \frac{27}{2}) : - \frac{63}{2}

= - \frac{63}{2}

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What is the integral from 0 to 3 of 7x(x-3) ?
The integral from 0 to 3 of 7x(x-3) is -63/2