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

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

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

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

1

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

9

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

Simplify

= 1

\int_{- π}^{0} - 1 d x

\int_{- 8}^{0} 64 - x^{2} d x

\int_{0}^{2 π} x^{2} + 1 d x

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

\int_{0}^{\frac{π}{2}} (sin (x))^{3} d x

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