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

The integral from 0 to x of (x^2)/9 is (x^3)/(27)

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

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

\frac{x ^{3}}{27}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{x ^{3}}{3}

= \frac{1}{9} \cdot \frac{x ^{3}}{3}

Simplify

= \frac{x ^{3}}{27}

\int_{1}^{e} \frac{9}{x} d x

\int_{0}^{1} \frac{1}{2} (1 - x^{2})^{2} d x

\int_{0}^{7} \frac{3}{x + 9} d x

\int_{0}^{\frac{π}{2}} (cos (x) - 1) d x

\int_{- 1}^{1} (\frac{x ^{3} - 8}{x - 2}) d x

What is the integral from 0 to x of (x^2)/9 ?
The integral from 0 to x of (x^2)/9 is (x^3)/(27)