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

The integral from 0 to 1 of x^2sqrt(1+3x^3) is 14/27

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

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

\frac{14}{27}

Decimal

0.51851 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{u}{9} d u

= \int_{1}^{4} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{1}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{1}{9} \cdot \frac{14}{3}

Simplify

= \frac{14}{27}

\int_{- 2}^{3} (3 x^{4} - 4 x^{3} - x - 4) d x

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

\int_{4}^{9} x^{\frac{3}{2}} d x

\int_{0}^{e^{- z^{4}}} z d z

\int_{0}^{x^{2}} t^{\frac{3}{2}} d t

What is the integral from 0 to 1 of x^2sqrt(1+3x^3) ?
The integral from 0 to 1 of x^2sqrt(1+3x^3) is 14/27