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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.