4812. f'(1)=1/√3
f'(2)=√3
f'(3)=1
The first part of integration comes up to be 1/3
the second part comes up to be 1-√3
Now I=1/3+1-√3 = 4/3-√3
1. If the normal to the curve y= f(x) at x=0 be given by the equation 3x - y + 3 = 0 ,then the value of lim x→0 [x2/{ f(x2) 5 f(4x2) + 4f(7x2)}-1 ] is ????
2.The tangent to the graph of the function y=f(x) at the point with abscissa x=1 form an angle of 30° and at the point x=2 an angle of 60° and at the point x=3 an angle of 45° ..The value of ∫ (upper limit=3,lower limit=1) f'(x) f''(x) dx + ∫ (upper limit=3,lower limit=2) f''(x)dx is???????(f''(x) is suppose to be continuous)
-
UP 0 DOWN 0 0 2
2 Answers
481
