What is the integral from 2/3 to 10 of (11-x)^{1/2} ?

The integral from 2/3 to 10 of (11-x)^{1/2} is (62*31^{1/2})/(9*3^{1/2)}-2/3

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integral from 2/3 to 10 of (11-x)^{1/2}

\int_{\frac{2}{3}}^{10} (11 - x)^{\frac{1}{2}} d x

\frac{62 \cdot 3 1 ^{\frac{1}{2}}}{9 \cdot 3 ^{\frac{1}{2}}} - \frac{2}{3}

Decimal Notation

21.47801 \dots

Solution steps

\int_{\frac{2}{3}}^{10} (11 - x)^{\frac{1}{2}} d x

Apply u-substitution

: \int_{\frac{31}{3}}^{1} - u^{\frac{1}{2}} d u

= \int_{\frac{31}{3}}^{1} - u^{\frac{1}{2}} d u

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

= - \int_{1}^{\frac{31}{3}} - u^{\frac{1}{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{1}^{\frac{31}{3}} u^{\frac{1}{2}} d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{31}{3}}

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{31}{3}})

Simplify

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{31}{3}}

Compute the boundaries:

\frac{62 \cdot 3 1 ^{\frac{1}{2}}}{9 \cdot 3 ^{\frac{1}{2}}} - \frac{2}{3}

= \frac{62 \cdot 3 1 ^{\frac{1}{2}}}{9 \cdot 3 ^{\frac{1}{2}}} - \frac{2}{3}

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What is the integral from 2/3 to 10 of (11-x)^{1/2} ?
The integral from 2/3 to 10 of (11-x)^{1/2} is (62*31^{1/2})/(9*3^{1/2)}-2/3