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

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

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

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

\frac{9}{4}

Solution steps

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

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{3} x x^{2} d x

Simplify

x x^{2} : x^{3}

= \frac{1}{9} \cdot \int_{0}^{3} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{3}

= \frac{1}{9} [\frac{x ^{4}}{4}]_{0}^{3}

Compute the boundaries:

\frac{81}{4}

= \frac{1}{9} \cdot \frac{81}{4}

Simplify

= \frac{9}{4}

\int_{2}^{3} \frac{1}{x ^{4}} d x

\int_{2}^{4} x^{2} e^{x^{3} + 5} d x

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

\int_{2}^{y} x^{2} sin (x) d x

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

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