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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.