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Ex 7.9, 20 - Direct Integrate (x ex + sin pi x / 4) dx - Ex 7.9

Ex 7.9, 20 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 20 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.9, 20 - Chapter 7 Class 12 Integrals - Part 4

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Ex 7.9, 20 ∫_0^1β–’(π‘₯ 𝑒^π‘₯+sinβ‘γ€–πœ‹π‘₯/4γ€— ) 𝑑π‘₯ Let F(π‘₯)=∫1β–’(π‘₯𝑒^π‘₯+𝑠𝑖𝑛 πœ‹π‘₯/4)𝑑π‘₯ =∫1β–’γ€–π‘₯𝑒^π‘₯ 𝑑π‘₯+∫1β–’γ€–sin⁑(πœ‹π‘₯/4) 𝑑π‘₯γ€—γ€— Solving I1 and I2 separately Solving π‘°πŸ ∫1β–’γ€–π‘₯𝑒^π‘₯ 𝑑π‘₯γ€— =π‘₯∫1▒〖𝑒^π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’[(𝑑π‘₯/𝑑π‘₯) ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€—]𝑑π‘₯γ€— =π‘₯𝑒^π‘₯βˆ’βˆ«1β–’(1.𝑒^π‘₯ 𝑑π‘₯)𝑑π‘₯ =π‘₯𝑒^π‘₯βˆ’βˆ«1▒〖𝑒^π‘₯ 𝑑π‘₯γ€— =π‘₯𝑒^π‘₯βˆ’π‘’^π‘₯ =𝑒^π‘₯ (π‘₯βˆ’1) Solving by parts ∫1▒〖𝑓(π‘₯)𝑔(π‘₯)𝑑π‘₯=𝑓(π‘₯)∫1▒〖𝑔(π‘₯)𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯)𝑑π‘₯)𝑑π‘₯γ€—γ€— Let 𝑓(π‘₯)=π‘₯ and 𝑔(π‘₯)=𝑒^π‘₯ Solving I2 ∫1β–’γ€–sin⁑(πœ‹π‘₯/4) 𝑑π‘₯γ€— = 1/(πœ‹/4) (βˆ’cos⁑(πœ‹π‘₯/4) ) = (βˆ’4)/πœ‹ cos⁑(πœ‹π‘₯/4) Therefore, F(π‘₯)=∫1β–’γ€–π‘₯𝑒^π‘₯ 𝑑π‘₯+∫1▒〖𝑠𝑖𝑛 πœ‹/4 π‘₯ 𝑑π‘₯γ€—γ€— =𝑒^π‘₯ (π‘₯βˆ’1)βˆ’4/πœ‹ cos⁑(πœ‹π‘₯/4) Now, ∫_0^1β–’(π‘₯𝑒^π‘₯+𝑠𝑖𝑛 πœ‹π‘₯/4) 𝑑π‘₯=𝐹(1)βˆ’πΉ(0) =(𝑒^1 (1βˆ’1)βˆ’4/πœ‹ cos⁑((πœ‹ Γ— 1)/4) )βˆ’(𝑒^0 (0βˆ’1)+4/πœ‹ π‘π‘œπ‘ ((πœ‹ Γ— 0)/4)) =𝑒×0βˆ’4/πœ‹ π‘π‘œπ‘  πœ‹/4βˆ’1(βˆ’1)+4/πœ‹ cos⁑0 =(βˆ’4)/( πœ‹) π‘π‘œπ‘  πœ‹/4+1+4/πœ‹ =(βˆ’4)/( πœ‹) 1/√2+1+4/πœ‹ =(βˆ’2√2)/( πœ‹) +1+4/πœ‹ =𝟏+πŸ’/π…βˆ’(𝟐√𝟐)/𝝅

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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.