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Ex 7.8, 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 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

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

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo