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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.