What is the integral from 0 to 2 of sqrt(4-x) ?

The integral from 0 to 2 of sqrt(4-x) is (16-4sqrt(2))/3

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integral from 0 to 2 of sqrt(4-x)

\int_{0}^{2} 4 - x d x

\frac{16 - 4 2}{3}

Solution steps

\int_{0}^{2} 4 - x d x

Apply u-substitution

: \int_{4}^{2} - u d u

= \int_{4}^{2} - u d u

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

= - \int_{2}^{4} - u d u

Take the constant out:

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

= - (- \int_{2}^{4} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{16}{3} - \frac{4 2}{3}

= \frac{16}{3} - \frac{4 2}{3}

Simplify

= \frac{16 - 4 2}{3}

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What is the integral from 0 to 2 of sqrt(4-x) ?
The integral from 0 to 2 of sqrt(4-x) is (16-4sqrt(2))/3