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

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

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

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

23

Decimal Notation

3.46410 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{0} - \frac{1}{u} d u

= \int_{3}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{3} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{0}^{3} \frac{1}{u} d u)

Apply the Power Rule

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

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

Simplify

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

= [2 u]_{0}^{3}

Compute the boundaries:

23

= 23

\int \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{2 ^{x}}{ln ( 2 )} x d x

\frac{d y}{d x} = \frac{( y ^{2} + x y + x ^{2} )}{x ^{2}}

\int_{0}^{2} 2 d x

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