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

The integral of x^2*sqrt(1-x) is -2/7 (1-x)^{7/2}+4/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+C

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integral of x^2*sqrt(1-x)

\int x^{2} \cdot 1 - x d x

- \frac{2}{7} (1 - x)^{\frac{7}{2}} + \frac{4}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int - (- u + 1)^{2} u d u

= \int - (- u + 1)^{2} u d u

Take the constant out:

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

= - \int (- u + 1)^{2} u d u

Expand

(- u + 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

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

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - x

= - (\frac{2}{7} (1 - x)^{\frac{7}{2}} - \frac{4}{5} (1 - x)^{\frac{5}{2}} + \frac{2}{3} (1 - x)^{\frac{3}{2}})

Simplify

= - \frac{2}{7} (1 - x)^{\frac{7}{2}} + \frac{4}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{7} (1 - x)^{\frac{7}{2}} + \frac{4}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + C

\frac{1}{2 x} - \frac{8}{5 x ^{4}}

\int \frac{lo g _{e} ( x )}{x} d x

f (x) = ln (10 x)

e^{5 x + 5}

x \to - \infty lim (x^{2} - x)

What is the integral of x^2*sqrt(1-x) ?
The integral of x^2*sqrt(1-x) is -2/7 (1-x)^{7/2}+4/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+C