What is the integral from 0 to 2 of 9yy) ?

The integral from 0 to 2 of 9yy) is 24d

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integral from 0 to 2 of 9yy)

\int_{0}^{2} 9 y d (y)

24 d

Solution steps

\int_{0}^{2} 9 y d y d y

Take the constant out:

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

= 9 d \cdot \int_{0}^{2} yy d y

Simplify

yy : y^{2}

= 9 d \cdot \int_{0}^{2} y^{2} d y

Apply the Power Rule

: [\frac{y ^{3}}{3}]_{0}^{2}

= 9 d [\frac{y ^{3}}{3}]_{0}^{2}

Compute the boundaries:

\frac{8}{3}

= 9 d \frac{8}{3}

Simplify

= 24 d

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What is the integral from 0 to 2 of 9yy) ?
The integral from 0 to 2 of 9yy) is 24d