What is the integral from 0 to 5 of x^2sqrt(25-x^2) ?

The integral from 0 to 5 of x^2sqrt(25-x^2) is (625pi)/(16)

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integral from 0 to 5 of x^2sqrt(25-x^2)

\int_{0}^{5} x^{2} 25 - x^{2} d x

\frac{625 π}{16}

Decimal Notation

122.71846 \dots

Solution steps

\int_{0}^{5} x^{2} 25 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 625 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 625 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

= 625 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 625 \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

= 625 \cdot \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

= 625 \cdot \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

= 625 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

625 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{625 π}{16}

= \frac{625 π}{16}

\frac{\partial}{\partial x} (ln (\frac{y}{x}))

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

n = 1 \sum \infty \frac{3 ^{n}}{5 ^{n}}

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

f (x) = 3 + 4 x^{2} - 2 x^{3}, (2, 3)

What is the integral from 0 to 5 of x^2sqrt(25-x^2) ?
The integral from 0 to 5 of x^2sqrt(25-x^2) is (625pi)/(16)