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

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

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

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

\frac{52}{9}

Decimal Notation

5.77777 \dots

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{52}{3}

= \frac{1}{3} \cdot \frac{52}{3}

Simplify

= \frac{52}{9}

x \to - \frac{1}{2} lim (\frac{6 x - 1}{x})

\int x^{2} 2 x - x^{2} d x

f (x) = x^{3} - 5 x^{2}, at x = 2

\frac{d y}{d x} = \frac{x ^{2} + x y + y ^{2}}{x ^{2}}

\frac{\partial}{\partial z} (- 4 x yz)

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