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

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

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integral from 0 to 1 of 1/(sqrt(4-2x))

\int_{0}^{1} \frac{1}{4 - 2 x} d x

2 - 2

Decimal

0.58578 \dots

Solution steps

\int_{0}^{1} \frac{1}{4 - 2 x} d x

Apply u-substitution

= \int_{4}^{2} - \frac{1}{2 u} d u

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

= - \int_{2}^{4} - \frac{1}{2 u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{2}^{4} \frac{1}{u} d u)

Apply the Power Rule

= - (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{2}^{4})

Simplify

- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{2}^{4}) : \frac{1}{2} [2 u]_{2}^{4}

= \frac{1}{2} [2 u]_{2}^{4}

Compute the boundaries:

4 - 22

= \frac{1}{2} (4 - 22)

Simplify

= 2 - 2

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