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

The integral of sqrt(x^2-x^3) is 2/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+C

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

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

\frac{2}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + C

Solution steps

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

Simplify

x^{2} - x^{3} : x 1 - x

= \int x 1 - x d x

Apply u-substitution

: \int - (- u + 1) u d u

= \int - (- u + 1) u d u

Take the constant out:

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

= - \int (- u + 1) u d u

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

= - \int - u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - x

= - (- \frac{2}{5} (1 - x)^{\frac{5}{2}} + \frac{2}{3} (1 - x)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + C

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y = \frac{5}{x}, y = \frac{10}{x ^{2}}, x = 6

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\int \frac{( e ^{x} )}{2} d x

What is the integral of sqrt(x^2-x^3) ?
The integral of sqrt(x^2-x^3) is 2/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+C