What is the integral from 0 to 8 of sqrt(25-2x) ?

The integral from 0 to 8 of sqrt(25-2x) is 98/3

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integral from 0 to 8 of sqrt(25-2x)

\int_{0}^{8} 25 - 2 x d x

\frac{98}{3}

Decimal

32.66666 \dots

Solution steps

\int_{0}^{8} 25 - 2 x d x

Apply u-substitution

: \int_{25}^{9} - \frac{u}{2} d u

= \int_{25}^{9} - \frac{u}{2} d u

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

= - \int_{9}^{25} - \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_{9}^{25} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

Compute the boundaries:

\frac{196}{3}

= \frac{1}{2} \cdot \frac{196}{3}

Simplify

= \frac{98}{3}

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What is the integral from 0 to 8 of sqrt(25-2x) ?
The integral from 0 to 8 of sqrt(25-2x) is 98/3