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

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

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

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

\frac{52}{9}

Decimal Notation

5.77777 \dots

Solution steps

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

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}

\frac{d}{d x} (\frac{2}{1 + 4 x ^{2}})

\int \frac{x ^{2}}{36 + x ^{2}} d x

x^{3} - 5

\frac{3 s - 1}{s ^{2} - 1}

\frac{\partial}{\partial x} (\frac{( x + y )}{y + z})

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