What is the integral from 1 to 2 of xsqrt(4-x^2) ?

The integral from 1 to 2 of xsqrt(4-x^2) is square root of 3

Popular >

integral from 1 to 2 of xsqrt(4-x^2)

\int_{1}^{2} x 4 - x^{2} d x

3

Decimal

1.73205 \dots

Solution steps

\int_{1}^{2} x 4 - x^{2} d x

Apply u-substitution

: \int_{3}^{0} - \frac{u}{2} d u

= \int_{3}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{3} - \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}^{3} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

23

= \frac{1}{2} \cdot 23

Simplify

= 3

\int_{0}^{1} x^{5} arcsin (x) d x

\int_{2 x}^{x^{3}} e^{- t^{2}} d t

\int_{3}^{10} \frac{1}{x ^{2}} d x

\int_{0}^{1} sin^{2} (x) cos^{5} (x) d x

\int_{0}^{1} y^{\frac{3}{2}} d y

What is the integral from 1 to 2 of xsqrt(4-x^2) ?
The integral from 1 to 2 of xsqrt(4-x^2) is square root of 3