What is the integral from 0 to 9 of 2sqrt(y) ?

The integral from 0 to 9 of 2sqrt(y) is 36

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integral from 0 to 9 of 2sqrt(y)

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

36

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

18

= 2 \cdot 18

Simplify

= 36

\frac{d}{d θ} (sec (θ) tan (θ))

\frac{3}{s - 9}

\frac{d}{d x} (f (x) g (x))

\frac{y}{y + \frac{a}{y}}

n = 0 \sum \infty \frac{n}{e ^{2 n}}

What is the integral from 0 to 9 of 2sqrt(y) ?
The integral from 0 to 9 of 2sqrt(y) is 36