What is the integral from 0 to 1 of (15)/(4y-1) ?

The integral from 0 to 1 of (15)/(4y-1) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (15)/(4y-1)

\int_{0}^{1} \frac{15}{4 y - 1} d y

diverges

Solution steps

\int_{0}^{1} \frac{15}{4 y - 1} d y

Find undefined (singularity) points:

y = \frac{1}{4}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{1}{4}} \frac{15}{4 y - 1} d y + \int_{\frac{1}{4}}^{1} \frac{15}{4 y - 1} d y

\int_{0}^{\frac{1}{4}} \frac{15}{4 y - 1} d y =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (x^{- l n (x)})

\int sin^{2} (2 t) cos (2 t) d t

x \to 5 lim (x 41 - x^{2})

\frac{d}{d v} (u^{2} + v^{2})

y^{^{''''}} = y

What is the integral from 0 to 1 of (15)/(4y-1) ?
The integral from 0 to 1 of (15)/(4y-1) is diverges