What is the integral from 0 to 2 of sqrt(1-0.5x) ?

The integral from 0 to 2 of sqrt(1-0.5x) is 4/3

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integral from 0 to 2 of sqrt(1-0.5x)

\int_{0}^{2} 1 - 0.5 x d x

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{0}^{2} 1 - 0.5 x d x

Apply u-substitution

: \int_{1}^{0} - \frac{u}{0.5} d u

= \int_{1}^{0} - \frac{u}{0.5} d u

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

= - \int_{0}^{1} - \frac{u}{0.5} d u

Take the constant out:

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

= - (- \frac{1}{0.5} \cdot \int_{0}^{1} u d u)

Simplify

= - (- 2 \cdot \int_{0}^{1} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{2}{3}

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

Simplify

= \frac{4}{3}

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What is the integral from 0 to 2 of sqrt(1-0.5x) ?
The integral from 0 to 2 of sqrt(1-0.5x) is 4/3