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

The integral from 0 to 3 of xsqrt(16-x^2) is (64-7sqrt(7))/3

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integral from 0 to 3 of xsqrt(16-x^2)

\int_{0}^{3} x 16 - x^{2} d x

\frac{64 - 7 7}{3}

Decimal Notation

15.15991 \dots

Solution steps

\int_{0}^{3} x 16 - x^{2} d x

Apply u-substitution

: \int_{16}^{7} - \frac{u}{2} d u

= \int_{16}^{7} - \frac{u}{2} d u

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

= - \int_{7}^{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_{7}^{16} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{16}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{16})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{16}

Compute the boundaries:

\frac{128}{3} - \frac{14 7}{3}

= \frac{1}{2} (\frac{128}{3} - \frac{14 7}{3})

Simplify

= \frac{64 - 7 7}{3}

\int \frac{1}{x ^{2} x ^{2} - 3} d x

x \to 0 lim (- x ln (x) - x)

\int x \cdot x d x

\frac{1}{x - 1}

x \cdot y^{^{'}} = x^{2} - 2 x y - y^{2} + y

What is the integral from 0 to 3 of xsqrt(16-x^2) ?
The integral from 0 to 3 of xsqrt(16-x^2) is (64-7sqrt(7))/3