What is the integral from 0 to 3 of (4x(9-x^2))/(81) ?

The integral from 0 to 3 of (4x(9-x^2))/(81) is 1

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integral from 0 to 3 of (4x(9-x^2))/(81)

\int_{0}^{3} \frac{4 x ( 9 - x ^{2} )}{81} d x

1

Solution steps

\int_{0}^{3} \frac{4 x ( 9 - x ^{2} )}{81} d x

Take the constant out:

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

= \frac{4}{81} \cdot \int_{0}^{3} x (9 - x^{2}) d x

Expand

x (9 - x^{2}) : 9 x - x^{3}

= \frac{4}{81} \cdot \int_{0}^{3} 9 x - x^{3} 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

= \frac{4}{81} (\int_{0}^{3} 9 x d x - \int_{0}^{3} x^{3} d x)

\int_{0}^{3} 9 x d x = \frac{81}{2}

\int_{0}^{3} x^{3} d x = \frac{81}{4}

= \frac{4}{81} (\frac{81}{2} - \frac{81}{4})

Simplify

\frac{4}{81} (\frac{81}{2} - \frac{81}{4}) : 1

= 1

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

\int_{0}^{π} x \cdot cos (2 x) d x

What is the integral from 0 to 3 of (4x(9-x^2))/(81) ?
The integral from 0 to 3 of (4x(9-x^2))/(81) is 1