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

The integral from 0 to a of x^2sqrt(a^2-x^2) is pi/(16)a^4

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

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

\frac{π}{16} a^{4}

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} a^{4} sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} a^{4} sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= a^{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= a^{4} \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= a^{4} \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 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

= a^{4} \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 u) d u = 0

= a^{4} \frac{1}{8} (\frac{π}{2} - 0)

Simplify

a^{4} \frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{16} a^{4}

= \frac{π}{16} a^{4}

\int_{1}^{e} 4 x ln (x) d x

(\frac{d y}{d x}) + 8 y = 12

\frac{d y}{d x} - y = 0

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

f (x) = 7 x^{15}

What is the integral from 0 to a of x^2sqrt(a^2-x^2) ?
The integral from 0 to a of x^2sqrt(a^2-x^2) is pi/(16)a^4