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  • Complete JEE Main/Advanced Course and Test Series
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Solved Examples on Area under Curves

Illustration 1: The area bounded by the curves y = f(x), x-axis and the ordinates x =1 and x = b is (b-1) sin (3b + 4). Then what is the value of f(x)?

Solution: We know that 

∫f(x) =  (b-1) sin (3b + 4), where integral runs from 1 to b

Now, differentiating both sides of the equation we get,

f(b) = 3(b-1) cos (3b + 4) + sin (3b + 4)

f(x) = sin (3x + 4) + 3(x-1) cos (3x + 4)


Illustration 2: Let the straight line x = b divided the area enclosed by y = (1-x)2, y = 0 and x = 0 into two parts R1 (0 ≤ x ≤ b) and R2 (b ≤ x ≤ 1) such that R1 – R2 = 1/4. Then what is the value of b? (2011)

Solution: It is given in the question that the area between 0 to b is R1 and b to 1 is R2.
Hence, -1/3 [(1-b)3 – 1] +1/3 [0 - (1-b)3] = 1/4
So, -2/3 (1-b)3 = -1/3 + 1/4 = -1/12
Hence, (1-b)3 = 1/8
So, 1-b =1/2 which gives b as 1/2.

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