What is the integral from 0 to 16 of 1/(64-x^2) ?

The integral from 0 to 16 of 1/(64-x^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 0 to 16 of 1/(64-x^2)

\int_{0}^{16} \frac{1}{64 - x ^{2}} d x

diverges

Solution steps

\int_{0}^{16} \frac{1}{64 - x ^{2}} d x

Find undefined (singularity) points:

x = 8

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}^{8} \frac{1}{64 - x ^{2}} d x + \int_{8}^{16} \frac{1}{64 - x ^{2}} d x

\int_{0}^{8} \frac{1}{64 - x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y^{3} - (14 x + 4) + 3 x y^{2} y^{^{'}} = 0

y + y^{4} = (y^{6} + 3 x) \frac{d y}{d x}

y = (\frac{( 3 x ^{2} )}{( 2 x + 4 )})^{3}

x \to 2 + lim (1)

\int x + x^{2} + 2 d x

What is the integral from 0 to 16 of 1/(64-x^2) ?
The integral from 0 to 16 of 1/(64-x^2) is diverges