What is the integral from 3 to 5 of 1/(sqrt(2x-6)) ?

The integral from 3 to 5 of 1/(sqrt(2x-6)) is 2

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integral from 3 to 5 of 1/(sqrt(2x-6))

\int_{3}^{5} \frac{1}{2 x - 6} d x

2

Solution steps

\int_{3}^{5} \frac{1}{2 x - 6} d x

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{2}

Simplify

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

Compute the boundaries:

2

= 2

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\int_{0}^{3} 3 x^{2} + 1 d x

\int_{0}^{1} e^{x} - e x d x

\int_{3}^{5} \frac{y - 3}{y - 2} d y

What is the integral from 3 to 5 of 1/(sqrt(2x-6)) ?
The integral from 3 to 5 of 1/(sqrt(2x-6)) is 2