What is the integral from 0 to 4 of sqrt(49-6x) ?

The integral from 0 to 4 of sqrt(49-6x) is 218/9

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integral from 0 to 4 of sqrt(49-6x)

\int_{0}^{4} 49 - 6 x d x

\frac{218}{9}

Solution steps

\int_{0}^{4} 49 - 6 x d x

Apply u-substitution

: \int_{49}^{25} - \frac{u}{6} d u

= \int_{49}^{25} - \frac{u}{6} d u

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

= - \int_{25}^{49} - \frac{u}{6} d u

Take the constant out:

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

= - (- \frac{1}{6} \cdot \int_{25}^{49} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{49}

= - (- \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{49})

Simplify

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{49}

Compute the boundaries:

\frac{436}{3}

= \frac{1}{6} \cdot \frac{436}{3}

Simplify

= \frac{218}{9}

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What is the integral from 0 to 4 of sqrt(49-6x) ?
The integral from 0 to 4 of sqrt(49-6x) is 218/9