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0-4pi-cosx-




Question Number 124352 by Mammadli last updated on 02/Dec/20
∫_0 ^(4π) ∥cosx∥=?
4π0cosx∥=?
Commented by mr W last updated on 02/Dec/20
do you mean ∫_0 ^(4π) ∣cosx∣ dx ?
doyoumean4π0cosxdx?
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
4π0cosxdx=8π20cosxdx=8[sinx]0π2=8(10)=8

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