Question 4
β«1_(π + π)^(π + π)βγ π(π₯) ππ₯γ is equal to
β«1_(π )^(π )βγπ(π₯βπ) ππ₯γ (B) β«1_(π )^(π )βγπ(π₯+π) ππ₯γ
(C) β«1_(π )^(π )βγπ(π₯) ππ₯γ (D) β«1_(π βπ)^(πβπ )βγπ(π₯) ππ₯γ
β«1_(π + π)^(π + π)βγ π(π₯) ππ₯γ
Putting π=π+π
Differentiating w.r.t. π₯
ππ₯=ππ‘
Now,
when π varies from a + c to b + c
then π varies from a to b
Therefore
β«1_(π + π)^(π + π)βγ π(π₯) ππ₯γ =β«_π^πβπ(π‘+π)ππ‘
Changing variables β using Property 1
=β«_π^πβπ(π+π)π π
So, the correct answer is (b)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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