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

The integral from 2 to 4 of xsqrt(16-x^2) is 8sqrt(3)

Popular >

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

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

83

Decimal

13.85640 \dots

Solution steps

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

Apply u-substitution

: \int_{12}^{0} - \frac{u}{2} d u

= \int_{12}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{12} - \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}^{12} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

Compute the boundaries:

163

= \frac{1}{2} \cdot 163

Simplify

= 83

\int_{- 1}^{3} 2 x + 3 d x

\int_{- 1}^{4} \frac{x}{x + 5} d x

\int_{- 12}^{12} (144 - x^{2}) d x

\int_{0}^{1} (x + e^{x}) d x

\int_{1}^{5} \frac{( x - 1 ) ^{3}}{x ^{2}} d x

What is the integral from 2 to 4 of xsqrt(16-x^2) ?
The integral from 2 to 4 of xsqrt(16-x^2) is 8sqrt(3)