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Solved Examples on Definite Integral


  1. 5/4                                                                                            2. 7
  3. 4                                                                                                4. 2

  Solution: Given that F(x) = ∫ f(t) dt, where integral runs from 0 to x.

  By Leibnitz rule, we have

  F’(x) = f(x)

  But F(x2) = x2 (1+x) = x2 + x3

  Hence, this gives F(x) = x + x3/2

  So, F’(x) = 1+ 3/2 x1/2

  Hence, f(x) = F’(x) = 1+ 3/2 x1/2

  Hence, f(4) = 1+ 3/2 . (4)1/2

  This gives f(4) = 1+ 3/2 . 2 = 4

  Hence, option (3) is correct.


  1. ±1                                                             2. ±1/ √2 

  3. ±1/ 2                                                         4. 0 and 1

  Solution: F(x) = ∫√2-t2 dt, where the integral runs from 1 to x

  Hence, f’(x) = √2-x2

  Another condition that is given in the question is that x2 – f’(x) = 0

  Therefore, x2 = √2-x2

  Hence, x4 = 2-x2

  So, x4 + x2 – 2 = 0

  This gives the value of x as ±1.
 

  1. f(1/2) < 1/2 and f(1/3) > 1/3

  2. f(1/2) > 1/2 and f(1/3) > 1/3

  3. f(1/2) < 1/2 and f(1/3) < 1/3

  4. f(1/2) > 1/2 and f(1/3) < 1/3

  Solution: The given condition is that ∫ √1- (f’(t))2 dt = ∫ f(t) dt , where integral runs from 0 to x and 0 ≤ x ≤ 1.

  Differentiating both sides with respect to x by using the Leibnitz rule, we...

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