What is the integral from 1 to 4 of 1/((x-2)^{3/2)} ?

The integral from 1 to 4 of 1/((x-2)^{3/2)} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of 1/((x-2)^{3/2)}

\int_{1}^{4} \frac{1}{( x - 2 ) ^{\frac{3}{2}}} d x

diverges

Solution steps

\int_{1}^{4} \frac{1}{( x - 2 ) ^{\frac{3}{2}}} d x

Adjust integral based on undefined pointsThere is an undefined point within the boundaries at

: 2

= \int_{1}^{2} \frac{1}{( x - 2 ) ^{\frac{3}{2}}} d x + \int_{2}^{4} \frac{1}{( x - 2 ) ^{\frac{3}{2}}} d x

\int_{2}^{4} \frac{1}{( x - 2 ) ^{\frac{3}{2}}} d x =

diverges

= diverges

\int_{0}^{1} x \cdot x d x

\int_{1}^{4} \frac{1}{( x + 3 ) ^{2}} d x

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

\int_{2}^{3} (2 + 2 y)^{2} d y

\int_{0}^{1} - 1 + \frac{π x}{2} d x

What is the integral from 1 to 4 of 1/((x-2)^{3/2)} ?
The integral from 1 to 4 of 1/((x-2)^{3/2)} is diverges