NIMCET 2020 Question Paper with solutions

Question : 27 of 120

Marks: +1, -0

I_n=∫↖{a}↙{0}(a^2-x^2)^ndx

where n is a positive integer, then the relation between lnand

I_{n-1}

Solution:

I_n=∫↖{a}↙{0}(a^2-x^2)^ndx

I_n

,put n = 2

I_2=∫↖{a}↙{0}(a^2-x^2)^2dx=∫↖{a}↙{0}a^4+x^4-2x^2a^2dx

[a^4x+x^5/5-2 x^3/3a^2]^a_0

a^5+a^x/5-{2a^5}/3⇒8/15a^5

I_{n-1}

put

n=2 \;I_1=1

I_1=∫↖{a}↙{0}(a^2-x^2)dx

[a^2x-x^3/3]^a_0⇒a^3-a^3/3⇒{2a^3}/3

By putting n=2 in option we get

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