What is the integral from 3 to 7 of (sqrt(4x-12))/2 ?

The integral from 3 to 7 of (sqrt(4x-12))/2 is 16/3

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integral from 3 to 7 of (sqrt(4x-12))/2

\int_{3}^{7} \frac{4 x - 12}{2} d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{3}^{7} \frac{4 x - 12}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{3}^{7} 4 x - 12 d x

Apply u-substitution

: \int_{0}^{16} \frac{u}{4} d u

= \frac{1}{2} \cdot \int_{0}^{16} \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int_{0}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= \frac{1}{2} \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Simplify

\frac{1}{2} \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16} : \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= \frac{1}{8} \cdot \frac{128}{3}

Simplify

= \frac{16}{3}

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What is the integral from 3 to 7 of (sqrt(4x-12))/2 ?
The integral from 3 to 7 of (sqrt(4x-12))/2 is 16/3