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

The integral from 0 to 4 of 2sqrt(4-x) is 32/3

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

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

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int_{4}^{0} - u d u

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

= 2 (- \int_{0}^{4} - u d u)

Take the constant out:

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

= 2 (- (- \int_{0}^{4} u d u))

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{16}{3}

= 2 \cdot \frac{16}{3}

Simplify

= \frac{32}{3}

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What is the integral from 0 to 4 of 2sqrt(4-x) ?
The integral from 0 to 4 of 2sqrt(4-x) is 32/3