What is the integral from 0 to 5 of sqrt((5-x)/5) ?

The integral from 0 to 5 of sqrt((5-x)/5) is 10/3

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integral from 0 to 5 of sqrt((5-x)/5)

\int_{0}^{5} \frac{5 - x}{5} d x

\frac{10}{3}

Decimal Notation

3.33333 \dots

Solution steps

\int_{0}^{5} \frac{5 - x}{5} d x

\frac{5 - x}{5} = \frac{1}{5} 5 - x

= \int_{0}^{5} \frac{1}{5} 5 - x d x

Take the constant out:

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

= \frac{1}{5} \cdot \int_{0}^{5} 5 - x d x

Apply u-substitution

: \int_{5}^{0} - u d u

= \frac{1}{5} \cdot \int_{5}^{0} - u d u

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

= \frac{1}{5} (- \int_{0}^{5} - u d u)

Take the constant out:

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

= \frac{1}{5} (- (- \int_{0}^{5} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{5}

= \frac{1}{5} (- (- [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{5}))

Simplify

= \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{5}

Compute the boundaries:

\frac{10 5}{3}

= \frac{1}{5} \frac{10 5}{3}

Simplify

= \frac{10}{3}

\frac{d}{d x} (\frac{x ^{3} - 1}{x ^{3} + 1})

f (x) = \frac{x cos ( x ) + π}{x - π}

\int 6 tan^{3} (2 x) sec^{5} (2 x) d x

\int \frac{1}{5 + 3 x} d x

What is the integral from 0 to 5 of sqrt((5-x)/5) ?
The integral from 0 to 5 of sqrt((5-x)/5) is 10/3