Definite Integration Question 47

Question 47

  1. If I=k=198kk+1k+1x(x+1)dx, then (a) I>loge99 (b) I<loge99 (c) I<4950 (d) I>4950

(2017 Adv.)

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Solution:

  1. I=k=198kk+1(k+1)x(x+1)dx

Clearly, I>k=198kk+1(k+1)(x+1)2dx

I>k=198(k+1)kk+11(x+1)2dx

I>k=198((k+1))1k+21k+1I>k=1981k+2

I>13++1100>98100I>4950

Also, I<k=198kk+1k+1x(k+1)dx=k=198[loge(k+1)logek] I<loge99



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