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Misc 35 - Prove definite integral 0->1 x ex dx = 1 - Miscellaneous

Misc 35 - Chapter 7 Class 12 Integrals - Part 2


Misc 35 Prove that ∫_0^1▒𝑥 𝑒^𝑥 𝑑𝑥=1 Solving L.H.S ∫_0^1▒𝑥 𝑒^𝑥 𝑑𝑥 First we will solve ∫1▒𝒙 𝒆^𝒙 𝒅𝒙 ∫1▒𝑥 𝑒^𝑥 𝑑𝑥 Now we know that ∫1▒〖𝑓(𝑥) 𝑔⁡(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓^′ (𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting value of f(x) = x and g(x) = ex ∫1▒𝑥 𝑒^𝑥 𝑑𝑥=𝑥∫1▒〖𝑒^𝑥 𝑑𝑥〗−∫1▒(𝑑𝑥/𝑑𝑥 ∫1▒〖𝑒^𝑥 𝑑𝑥〗) 𝑑𝑥 = 𝑥𝑒^𝑥−∫1▒1. 𝑒^𝑥 𝑑𝑥 = 𝑥𝑒^𝑥−𝑒^𝑥+𝐶 Applying limits ∫1_0^1▒〖𝑥 𝑒^𝑥 𝑑𝑥〗 = [𝑥𝑒^𝑥−𝑒^𝑥 ]_0^1 = (1𝑒^1−𝑒^1 )−(0.𝑒^0−𝑒^0) = (𝑒^1−𝑒^1 )− (0 − 1) = 1 = R.H.S Hence proved.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.