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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Apply u-substitution

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

= \int_{4}^{0} - \frac{u}{4} d u

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

= - \int_{0}^{4} - \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}{4} \cdot \int_{0}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})

Simplify

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

Compute the boundaries:

\frac{16}{3}

= \frac{1}{4} \cdot \frac{16}{3}

Simplify

= \frac{4}{3}

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