Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 124352 by Mammadli last updated on 02/Dec/20

∫_0 ^(4π) ∥cosx∥=?

$$\underset{\mathrm{0}} {\overset{\mathrm{4}\pi} {\int}}\parallel{cosx}\parallel=? \\ $$

Commented by mr W last updated on 02/Dec/20

do you mean ∫_0 ^(4π) ∣cosx∣ dx ?

$${do}\:{you}\:{mean}\:\underset{\mathrm{0}} {\overset{\mathrm{4}\pi} {\int}}\mid{cosx}\mid\:{dx}\:? \\ $$

Commented by Mammadli last updated on 02/Dec/20

Dear Ser, as I mentioned, it was given..

Commented by mr W last updated on 02/Dec/20

∫_0 ^(4π) ∣cosx∣ dx   =8 ∫_0 ^(π/2) cos x dx   =8[sin x]_0 ^(π/2)   =8(1−0)  =8

$$\underset{\mathrm{0}} {\overset{\mathrm{4}\pi} {\int}}\mid{cosx}\mid\:{dx}\: \\ $$$$=\mathrm{8}\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{cos}\:{x}\:{dx}\: \\ $$$$=\mathrm{8}\left[\mathrm{sin}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{8}\left(\mathrm{1}−\mathrm{0}\right) \\ $$$$=\mathrm{8} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com