What is the integral of 1/2 x^3sqrt(4-x^2) ?

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

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integral of 1/2 x^3sqrt(4-x^2)

\int \frac{1}{2} x^{3} 4 - x^{2} d x

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

Solution steps

\int \frac{1}{2} x^{3} 4 - x^{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{3} 4 - x^{2} d x

Apply u-substitution

: \int - (- u^{2} + 4) u^{2} d u

= \frac{1}{2} \cdot \int - (- u^{2} + 4) u^{2} d u

Take the constant out:

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

= \frac{1}{2} (- \int (- u^{2} + 4) u^{2} d u)

Expand

(- u^{2} + 4) u^{2} : - u^{4} + 4 u^{2}

= \frac{1}{2} (- \int - u^{4} + 4 u^{2} 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

= \frac{1}{2} (- (- \int u^{4} d u + \int 4 u^{2} d u))

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

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

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

Substitute back

u = 4 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{1}{x ^{8}}

\int_{0}^{3} x e^{x} d x

\int_{0}^{6} (6 - x)^{2} d x

\int_{2}^{4} x d x

2 x^{2}, 2 x + 6, 1, 2

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