24sir its OK
I have done it.
A simialr ques on area was posted a few days back..here is another one
f(x)=\sin x
g(x)=\begin{Bmatrix} (max f(t),0\leq t\leq x) &0\leq x\leq \pi \\ (1-\cos x )/2& x> \pi \end{Bmatrix}
Discuss continuity and differentiablity of g(x) in (0,∞).Also find area enclosed
-
UP 0 DOWN 0 1 3
3 Answers
62g(x) = sin x (0<=x<pi/2)
= 1 (pi/2<=x<=pi)
= (1-cos x)/2 x>pi
Now you can solve it?
62I am giving the answer mentally so I may be wrong...
pi/2 - 1/2 (for the region between pi/2 and pi)
+ pi/2 + 3 (for the region between pi and 2 pi)
+ some area which i cant mentally from 2pi to the point of intersection of sin x and (1-cos x)/2 just after 2pi...