What is the integral from 0 to 7 of x^2sqrt(49-x^2) ?

The integral from 0 to 7 of x^2sqrt(49-x^2) is (2401pi)/(16)

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integral from 0 to 7 of x^2sqrt(49-x^2)

\int_{0}^{7} x^{2} 49 - x^{2} d x

\frac{2401 π}{16}

Decimal Notation

471.43524 \dots

Solution steps

\int_{0}^{7} x^{2} 49 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 2401 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 2401 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

= 2401 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 2401 \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

= 2401 \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

= 2401 \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

= 2401 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

2401 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{2401 π}{16}

= \frac{2401 π}{16}

y^{^{'}} = (2 t + 1) e^{y}

\frac{\partial}{\partial y} (ln (x^{4} y))

\frac{d}{d x} ((2 - 3 x^{2})^{4} (x^{7} + 3)^{3})

\frac{d y}{d x} + 7 y = 4

\frac{d}{d z} (z + 1 - 1)

What is the integral from 0 to 7 of x^2sqrt(49-x^2) ?
The integral from 0 to 7 of x^2sqrt(49-x^2) is (2401pi)/(16)