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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

Apply u-substitution

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

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

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

Take the constant out:

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

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

Apply the Power Rule

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

Simplify

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

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

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