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

The integral of sqrt(3-x)*x^2 is -2/7 (3-x)^{7/2}+12/5 (3-x)^{5/2}-6(3-x)^{3/2}+C

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

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

- \frac{2}{7} (3 - x)^{\frac{7}{2}} + \frac{12}{5} (3 - x)^{\frac{5}{2}} - 6 (3 - x)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

(- u + 3)^{2} u : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= - \int u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 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 6 u^{\frac{3}{2}} d u + \int 9 u d u)

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

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 3 - x

= - (\frac{2}{7} (3 - x)^{\frac{7}{2}} - \frac{12}{5} (3 - x)^{\frac{5}{2}} + 6 (3 - x)^{\frac{3}{2}})

Simplify

= - \frac{2}{7} (3 - x)^{\frac{7}{2}} + \frac{12}{5} (3 - x)^{\frac{5}{2}} - 6 (3 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{7} (3 - x)^{\frac{7}{2}} + \frac{12}{5} (3 - x)^{\frac{5}{2}} - 6 (3 - x)^{\frac{3}{2}} + C

\int \frac{3 x 4 x ^{2} - 6}{5} d x

\int_{1}^{\infty} \frac{y}{y ^{4} + 1} d y

e^{9 x}

f (t) = sin^{2} (t)

\frac{x ^{3}}{2}

What is the integral of sqrt(3-x)*x^2 ?
The integral of sqrt(3-x)*x^2 is -2/7 (3-x)^{7/2}+12/5 (3-x)^{5/2}-6(3-x)^{3/2}+C