Area under the curveHard
Question
If y = f(x) makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then 
xf′(x) dx is
xf′(x) dx isOptions
A.3/2
B.1
C.5/4
D.-3/4
Solution
We have f(x) dx = 
; Now, xf′(x)dx = x f′(x)dx - f(x) dx
= [x f(x)]02 - = 2f(2)- = 0 - (∵f(2) = 0) = -
f(x) dx = ; Now, xf′(x)dx = x f′(x)dx - f(x) dx= [x f(x)]02 -
= 2f(2)- = 0 - (∵f(2) = 0) = - Create a free account to view solution
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