1b/w ans wrong hai it's C,D
consider the integral \int_{0}^{inf}{x^{\mu}}dx where μ is real . We have \int_{0}^{inf}{x^{\mu}}dx = \int_{0}^{1}{x^{\mu}}dx + \int_{1}^{inf}{x^{\mu}}dx = I1 + I2
, where I1 is evaluated using the limitng process \lim_{\epsilon \rightarrow 0}\int_{0}^{1}{x^{\mu }} d\mu
and I2 is calculated using I_{2} = \int_{1}^{inf}{x^{\mu }}dx is evaluated using the limitng process \lim_{R\rightarrow inf}\int_{1}^{R}{x^{\mu }} d\mu
it follows that .
A) both integrals exist for all μ+1>0
B) both integrals exsit for all μ+1 < 0
C) I1 exists and I2 doesn't exist if μ+1>0
D) I1 doesn't exist and I2 exist if μ+1 < 0
it's multi correct options .
how do I proceed ? what significance does μ+1 have ?????? mughe kuch nahi samaj me nahi aaya isme :(
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5 Answers
1please post ur mathematics doubts here
http://targetiit.com/iit_jee_forum/posts/mathematics_4452.html
66I_1=\lim_{\epsilon\to0}\int_\epsilon^1x^\mu\ \mathrm{d}x=\lim_{\epsilon\to0} \dfrac{1-\epsilon^{\,\mu+1}}{\mu+1}
For I1 to exist, it is quite obvious that \mu+1>0, otherwise we shall be dividing by 0.
On the other hand,
I_2=\lim_{R\to\infty}\int_1^Rx^\mu \ \mathrm{d}x=\lim_{R\to\infty}\dfrac{R^{\mu +1}-1}{\mu +1}
For I2 to exist, we must have \mu +1<0, or else the limit would be infinite.
That is to say that for \mu+1>0, I1 exists but I2 does not, while for \mu +1<0, I2 exists but not I1.
So the options (C) and (D) are correct.