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

The integral from 3 to 4 of 3/(sqrt(x-3)) is 6

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integral from 3 to 4 of 3/(sqrt(x-3))

\int_{3}^{4} \frac{3}{x - 3} d x

6

Solution steps

\int_{3}^{4} \frac{3}{x - 3} d x

Take the constant out:

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

= 3 \cdot \int_{3}^{4} \frac{1}{x - 3} d x

Apply u-substitution

: \int_{0}^{1} 2 d u

= 3 \cdot \int_{0}^{1} 2 d u

Integral of a constant:

\int a d x = a x

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

Compute the boundaries:

2

= 3 \cdot 2

Simplify

= 6

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What is the integral from 3 to 4 of 3/(sqrt(x-3)) ?
The integral from 3 to 4 of 3/(sqrt(x-3)) is 6