What is the integral from 0 to a of x^3sqrt(a^2-x^2) ?

The integral from 0 to a of x^3sqrt(a^2-x^2) is (2a^5)/(15)

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integral from 0 to a of x^3sqrt(a^2-x^2)

\int_{0}^{a} x^{3} a^{2} - x^{2} d x

\frac{2 a ^{5}}{15}

Solution steps

\int_{0}^{a} x^{3} a^{2} - x^{2} d x

Apply u-substitution

= \int_{a^{2}}^{0} - \frac{( - u + a ^{2} ) 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} \cdot \int_{a^{2}}^{0} (- u + a^{2}) u d u

Expand

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

= - \frac{1}{2} \cdot \int_{a^{2}}^{0} - u^{\frac{3}{2}} + a^{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

= - \frac{1}{2} (- \int_{a^{2}}^{0} u^{\frac{3}{2}} d u + \int_{a^{2}}^{0} a^{2} u d u)

\int_{a^{2}}^{0} u^{\frac{3}{2}} d u = - \frac{2 ( a ^{2} ) ^{\frac{5}{2}}}{5}

\int_{a^{2}}^{0} a^{2} u d u = - \frac{2 a ^{2} ( a ^{2} ) ^{\frac{3}{2}}}{3}

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

Simplify

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

= \frac{2 a ^{5}}{15}

\int_{1}^{2} 2 x (x^{2} + 5) d x

\int_{0}^{1} 2^{2 x + 1} d x

\int_{0}^{4} (x x^{2} + 9) d x

\int_{0}^{5} (3 t - 8) d t

\int_{2}^{3} (3 x + 8) d x

What is the integral from 0 to a of x^3sqrt(a^2-x^2) ?
The integral from 0 to a of x^3sqrt(a^2-x^2) is (2a^5)/(15)