What is the integral from 0 to 2 of sqrt(25-3x) ?

The integral from 0 to 2 of sqrt(25-3x) is (250-38sqrt(19))/9

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integral from 0 to 2 of sqrt(25-3x)

\int_{0}^{2} 25 - 3 x d x

\frac{250 - 38 19}{9}

Decimal Notation

9.37353 \dots

Solution steps

\int_{0}^{2} 25 - 3 x d x

Apply u-substitution

: \int_{25}^{19} - \frac{u}{3} d u

= \int_{25}^{19} - \frac{u}{3} d u

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

= - \int_{19}^{25} - \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_{19}^{25} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{19}^{25}

= - (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{19}^{25})

Simplify

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{19}^{25}

Compute the boundaries:

\frac{250}{3} - \frac{38 19}{3}

= \frac{1}{3} (\frac{250}{3} - \frac{38 19}{3})

Simplify

= \frac{250 - 38 19}{9}

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What is the integral from 0 to 2 of sqrt(25-3x) ?
The integral from 0 to 2 of sqrt(25-3x) is (250-38sqrt(19))/9