What is the integral from 1/3 to 2 of sqrt(10-3p) ?

The integral from 1/3 to 2 of sqrt(10-3p) is 38/9

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integral from 1/3 to 2 of sqrt(10-3p)

\int_{\frac{1}{3}}^{2} 10 - 3 p d p

\frac{38}{9}

Decimal

4.22222 \dots

Solution steps

\int_{\frac{1}{3}}^{2} 10 - 3 p d p

Apply u-substitution

: \int_{9}^{4} - \frac{u}{3} d u

= \int_{9}^{4} - \frac{u}{3} d u

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

= - \int_{4}^{9} - \frac{u}{3} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{4}^{9} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

= - (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9})

Simplify

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

Compute the boundaries:

\frac{38}{3}

= \frac{1}{3} \cdot \frac{38}{3}

Simplify

= \frac{38}{9}

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What is the integral from 1/3 to 2 of sqrt(10-3p) ?
The integral from 1/3 to 2 of sqrt(10-3p) is 38/9