What is the integral from 0 to 2 of 9y^2sqrt(y^3+1) ?

The integral from 0 to 2 of 9y^2sqrt(y^3+1) is 52

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integral from 0 to 2 of 9y^2sqrt(y^3+1)

\int_{0}^{2} 9 y^{2} y^{3} + 1 d y

52

Solution steps

\int_{0}^{2} 9 y^{2} y^{3} + 1 d y

Take the constant out:

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

= 9 \cdot \int_{0}^{2} y^{2} y^{3} + 1 d y

Apply u-substitution

: \int_{1}^{9} \frac{u}{3} d u

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= 9 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Simplify

9 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9} : 3 [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= 3 [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= 3 \cdot \frac{52}{3}

Simplify

= 52

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

\int_{0}^{3} (x - 3) e^{- x} d x

\int_{0}^{3} x 5^{x^{2}} d x

\int_{0}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

\int_{1}^{1} 1

What is the integral from 0 to 2 of 9y^2sqrt(y^3+1) ?
The integral from 0 to 2 of 9y^2sqrt(y^3+1) is 52