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

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

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

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

\frac{52}{9}

Decimal

5.77777 \dots

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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})

Simplify

= \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}

\int_{0}^{a r c t a n (x)} sec^{2} (t) d t

\int_{2}^{3} (4 x^{3} - \frac{x ^{3}}{6}) d x

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

\int_{0}^{\infty} e^{- y} (2 y^{3} - \frac{2 y ^{3}}{3}) d y

\int_{1}^{9} \frac{11}{x} d x

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