What is the integral of 2/3 (1-x)^{3/2}(1) ?

The integral of 2/3 (1-x)^{3/2}(1) is -4/15 (1-x)^{5/2}+C

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integral of 2/3 (1-x)^{3/2}(1)

\int \frac{2}{3} (1 - x)^{\frac{3}{2}} (1) d x

- \frac{4}{15} (1 - x)^{\frac{5}{2}} + C

Solution steps

\int \frac{2}{3} (1 - x)^{\frac{3}{2}} \cdot 1 d x

Take the constant out:

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

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

Simplify

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

Apply u-substitution

: \int - u^{\frac{3}{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

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

Substitute back

u = 1 - x

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

Simplify

\frac{2}{3} (- \frac{2}{5} (1 - x)^{\frac{5}{2}}) : - \frac{4}{15} (1 - x)^{\frac{5}{2}}

= - \frac{4}{15} (1 - x)^{\frac{5}{2}}

Add a constant to the solution

= - \frac{4}{15} (1 - x)^{\frac{5}{2}} + C

\frac{d}{d x} (\frac{6 x}{5 - tan ( x )})

y^{^{''}} - 3 y^{^{'}} + 2 = e^{t}

x \to \infty lim (2 + x)

\int_{0}^{1} \frac{1}{x ln ( x )} d x

y = e^{3 e^{x}}

What is the integral of 2/3 (1-x)^{3/2}(1) ?
The integral of 2/3 (1-x)^{3/2}(1) is -4/15 (1-x)^{5/2}+C