What is the integral of 7x^2sqrt(1-x^3) ?

The integral of 7x^2sqrt(1-x^3) is -14/9 (1-x^3)^{3/2}+C

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integral of 7x^2sqrt(1-x^3)

\int 7 x^{2} 1 - x^{3} d x

- \frac{14}{9} (1 - x^{3})^{\frac{3}{2}} + C

Solution steps

\int 7 x^{2} 1 - x^{3} d x

Take the constant out:

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

= 7 \cdot \int x^{2} 1 - x^{3} d x

Apply u-substitution

: \int - \frac{u}{3} d u

= 7 \cdot \int - \frac{u}{3} d u

Take the constant out:

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

= 7 (- \frac{1}{3} \cdot \int u d u)

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 7 (- \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - x^{3}

= 7 (- \frac{1}{3} \cdot \frac{2}{3} (1 - x^{3})^{\frac{3}{2}})

Simplify

7 (- \frac{1}{3} \cdot \frac{2}{3} (1 - x^{3})^{\frac{3}{2}}) : - \frac{14}{9} (1 - x^{3})^{\frac{3}{2}}

= - \frac{14}{9} (1 - x^{3})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{14}{9} (1 - x^{3})^{\frac{3}{2}} + C

\frac{d y}{e ^{5 y}} = e^{4 x} d x

x \to \infty lim (- x^{4})

\frac{d y}{d x} = \frac{x ^{5} - 2 y}{x}

f (x) = e^{x^{3} + 1}

What is the integral of 7x^2sqrt(1-x^3) ?
The integral of 7x^2sqrt(1-x^3) is -14/9 (1-x^3)^{3/2}+C