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

The integral from 0 to 4 of xsqrt(16-x^2) is 64/3

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

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

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{16}^{0} - \frac{u}{2} d u

= \int_{16}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{16} - \frac{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_{0}^{16} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= \frac{1}{2} \cdot \frac{128}{3}

Simplify

= \frac{64}{3}

\int \frac{1 - t}{t} d t

x \to 1 lim (3 x^{3} - 5 x^{2} + x + 5)

\frac{d}{d x} (- \frac{1}{5} cos (5 x))

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 3

y y^{^{'}} - 6 e^{x} = 0

What is the integral from 0 to 4 of xsqrt(16-x^2) ?
The integral from 0 to 4 of xsqrt(16-x^2) is 64/3