The area under rcosθ =1 + tanθ, θ = -π/4 to θ = 0, is the same as the area under y=x(x-1), x = 0 to x=1.State true or false?Give reason?

Learning is just like eating , it is not how much we eat it is how much we can digest.

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- Aug 22nd 2008, 11:37 PMpuneetarea under curve
The area under rcosθ =1 + tanθ, θ = -π/4 to θ = 0, is the same as the area under y=x(x-1), x = 0 to x=1.State true or false?Give reason?

Learning is just like eating , it is not how much we eat it is how much we can digest.

- Aug 23rd 2008, 12:05 AMkalagota

see.. if we set $\displaystyle x = r\cos \theta$ and $\displaystyle y = r\sin \theta$

$\displaystyle \tan \theta = \frac{y}{x}$

thus, $\displaystyle r\cos\theta = 1 + \tan \theta \Longleftrightarrow x = 1 + \frac{y}{x} \Longleftrightarrow y = x(x-1)$

how about the limits of integration?

try to see if they give you same limits.. - Aug 23rd 2008, 02:35 AMpuneetthanks
thanks