What is the integral from 0 to 6 of (16-2x)^{1/2} ?

The integral from 0 to 6 of (16-2x)^{1/2} is 56/3

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integral from 0 to 6 of (16-2x)^{1/2}

\int_{0}^{6} (16 - 2 x)^{\frac{1}{2}} d x

\frac{56}{3}

Decimal Notation

18.66666 \dots

Solution steps

\int_{0}^{6} (16 - 2 x)^{\frac{1}{2}} d x

(16 - 2 x)^{\frac{1}{2}} = 2^{\frac{1}{2}} (8 - x)^{\frac{1}{2}}

= \int_{0}^{6} 2^{\frac{1}{2}} (8 - x)^{\frac{1}{2}} d x

Take the constant out:

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

= 2^{\frac{1}{2}} \cdot \int_{0}^{6} (8 - x)^{\frac{1}{2}} d x

Apply u-substitution

: \int_{8}^{2} - u^{\frac{1}{2}} d u

= 2^{\frac{1}{2}} \cdot \int_{8}^{2} - u^{\frac{1}{2}} d u

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

= 2^{\frac{1}{2}} (- \int_{2}^{8} - u^{\frac{1}{2}} d u)

Take the constant out:

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

= 2^{\frac{1}{2}} (- (- \int_{2}^{8} u^{\frac{1}{2}} d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{8}

= 2^{\frac{1}{2}} (- (- [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{8}))

Simplify

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

Compute the boundaries:

\frac{28 \cdot 2 ^{\frac{1}{2}}}{3}

= 2^{\frac{1}{2}} \cdot \frac{28 \cdot 2 ^{\frac{1}{2}}}{3}

Simplify

= \frac{56}{3}

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What is the integral from 0 to 6 of (16-2x)^{1/2} ?
The integral from 0 to 6 of (16-2x)^{1/2} is 56/3