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

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

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integral from 0 to 4 of x^2sqrt(16-x^2)

\int_{0}^{4} x^{2} 16 - x^{2} d x

16 π

Decimal Notation

50.26548 \dots

Solution steps

\int_{0}^{4} x^{2} 16 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 256 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 256 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

= 256 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 256 \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

= 256 \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

= 256 \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

= 256 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

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

= 16 π

\frac{d y}{d x} - x y^{2} = 2 x y

n = 0 \sum \infty \frac{1}{2 ^{2 n}}

\int \frac{1}{6} e^{6 x} d x

x \to 3 lim (x^{2} - 3 x - 10)

\int \frac{x ^{2} - 16 x + 9}{x ^{4} + 10 x ^{2} + 9} d x

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