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

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

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integral from 0 to 1 of 1/(sqrt(3-2x))

\int_{0}^{1} \frac{1}{3 - 2 x} d x

3 - 1

Decimal Notation

0.73205 \dots

Solution steps

\int_{0}^{1} \frac{1}{3 - 2 x} d x

Apply u-substitution

: \int_{3}^{1} - \frac{1}{2 u} d u

= \int_{3}^{1} - \frac{1}{2 u} d u

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

= - \int_{1}^{3} - \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_{1}^{3} \frac{1}{u} d u)

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

23 - 2

= \frac{1}{2} (23 - 2)

Simplify

= 3 - 1

f (x) = x^{2}, at x = 4

\frac{d}{d x} (2 + 3 e^{- \frac{3}{x}})

\frac{d}{d x} ((x^{4} + 3 x^{2} - 5)^{4})

\frac{d}{d x} (3 x e^{5 x})

\frac{d}{d x} (5^{- \frac{1}{x}})

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