What is the integral from 4 to 13 of 1/(sqrt(x-4)) ?

The integral from 4 to 13 of 1/(sqrt(x-4)) is 6

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integral from 4 to 13 of 1/(sqrt(x-4))

\int_{4}^{13} \frac{1}{x - 4} d x

6

Solution steps

\int_{4}^{13} \frac{1}{x - 4} d x

Apply u-substitution

: \int_{0}^{3} 2 d u

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

Integral of a constant:

\int a d x = a x

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

Compute the boundaries:

6

= 6

\int_{3}^{4} (x - 4)^{3} - (x - 4) d x

\int_{r^{1}}^{r^{2}} \frac{1}{r ^{2}} d r

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

\int_{0}^{2} \frac{x e ^{x}}{( 1 + x ) ^{2}} d x

\int_{1}^{5} x^{2} - 4 x - 12 d x

What is the integral from 4 to 13 of 1/(sqrt(x-4)) ?
The integral from 4 to 13 of 1/(sqrt(x-4)) is 6