Integration Full Chapter Explained - Integration Class 12 - Everything you need

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

Transcript

Example 43 (Introduction) Evaluate ∫_(βˆ’1)^(3/2)β–’|π‘₯ sin⁑(πœ‹ π‘₯) | 𝑑π‘₯ To find sign of |π‘₯ sin⁑(πœ‹ π‘₯) | in the interval, let us check sign of x and sin⁑〖 (πœ‹π‘₯) γ€—separately π‘₯ > 0 & π‘₯ sin⁑〖 (πœ‹π‘₯) γ€—> 0 π‘₯ < 0 & π‘₯ sin⁑〖 (πœ‹π‘₯) γ€—< 0 𝒙 π’”π’Šπ’β‘γ€–(𝝅𝒙) γ€—< 0 π‘₯ < 0 & π‘₯ sin⁑〖 (πœ‹π‘₯) γ€—> 0 π‘₯ > 0 & π‘₯ sin⁑〖 (πœ‹π‘₯) γ€—< 0 Sign of x We have Interval βˆ’1< π‘₯ < 3/2 Sign of 𝐬𝐒𝐧⁑〖(𝝅𝒙)γ€— Here, βˆ’1≀π‘₯≀3/2 βˆ’πœ‹ β‰€πœ‹π‘₯≀3πœ‹/2 Let πœƒ=πœ‹π‘₯ ∴ βˆ’πœ‹ ≀θ≀3πœ‹/2 From graph it is seen, Thus, Sign for 𝐬𝐒𝐧⁑〖(𝝅𝒙)γ€— is Now, Sign for π‘₯ sin⁑(πœ‹π‘₯) in interval –1 ≀ x ≀ 3/2 is So, we can write |π‘₯ 𝑠𝑖𝑛 (Ο€π‘₯)|={β–ˆ(π‘₯ sin⁑〖 (πœ‹π‘₯)γ€— βˆ’1 ≀ π‘₯ ≀ 1@&βˆ’π‘₯ sin⁑〖 (πœ‹π‘₯)γ€— 1 ≀ π‘₯ ≀ 3/2)─ Example 43 Evaluate ∫_(βˆ’1)^(3/2)β–’|π‘₯ sin⁑(πœ‹ π‘₯) | 𝑑π‘₯ Hence, |π‘₯ 𝑠𝑖𝑛 (Ο€π‘₯)|={β–ˆ(π‘₯ sin⁑〖 (πœ‹π‘₯)γ€— βˆ’1 ≀ π‘₯ ≀ 1@&βˆ’π‘₯ sin⁑〖 (πœ‹π‘₯)γ€— 1 ≀ π‘₯ ≀ 3/2)─ Therefore, we can write ∴ ∫_(βˆ’1)^(3/2)β–’|π‘₯ sin⁑(πœ‹π‘₯) | 𝑑π‘₯ = ∫_(βˆ’1)^1β–’γ€–π‘₯ sin⁑〖 (πœ‹π‘₯) 𝑑π‘₯ βˆ’βˆ«_1^(3/2)β–’γ€–π‘₯ 𝑠𝑖𝑛〗〗 γ€—(πœ‹π‘₯) 𝑑π‘₯ Solving ∫1▒〖𝒙 π’”π’Šπ’β‘γ€– (𝝅𝒙) 𝒅𝒙〗 γ€— separately ∫1β–’γ€–π‘₯ sin⁑〖 (πœ‹π‘₯) 𝑑π‘₯γ€— γ€— = x ∫1β–’sin⁑〖(πœ‹π‘₯)βˆ’βˆ«1β–’(𝑑(π‘₯)/𝑑π‘₯ ∫1β–’sin⁑〖(πœ‹π‘₯) γ€— ) γ€— 𝑑π‘₯ = x ((βˆ’cos⁑〖(πœ‹π‘₯))γ€—)/πœ‹βˆ’βˆ«1β–’1((βˆ’cos⁑〖(πœ‹π‘₯)γ€—)/πœ‹) 𝑑π‘₯ = (βˆ’ π‘₯ cos⁑〖(πœ‹π‘₯) γ€—)/πœ‹+∫1β–’cos⁑〖(πœ‹π‘₯)γ€—/πœ‹ 𝑑π‘₯ = (βˆ’π‘₯ cos⁑〖(πœ‹π‘₯)γ€—)/πœ‹+ sin⁑〖(πœ‹π‘₯)γ€—/πœ‹^2 Using by parts ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = x & g(x) = sin (πœ‹x) Putting limits ∫_(βˆ’πŸ)^πŸβ–’γ€–π’™ 𝐬𝐒𝐧⁑〖𝝅𝒙 𝒅𝒙〗 γ€— = [(βˆ’π‘₯ cos⁑〖(πœ‹π‘₯)γ€—)/πœ‹+sin⁑〖(πœ‹π‘₯)γ€—/πœ‹^2 ]_(βˆ’1)^1 = ((βˆ’1 cosβ‘πœ‹)/πœ‹+sinβ‘πœ‹/πœ‹^2 ) βˆ’((βˆ’(βˆ’1)cos⁑〖(βˆ’πœ‹)γ€—)/πœ‹+sin⁑〖(βˆ’πœ‹)γ€—/πœ‹^2 ) = ((βˆ’1 Γ— (βˆ’1))/πœ‹+0/πœ‹^2 ) βˆ’(cosβ‘πœ‹/πœ‹+γ€–βˆ’ sinγ€—β‘πœ‹/πœ‹^2 ) = (1/πœ‹+0)βˆ’((βˆ’1)/πœ‹+0) = 1/πœ‹+1/πœ‹ = 2/πœ‹ Putting limits ∫_𝟏^(πŸ‘/𝟐)▒〖𝒙 𝐬𝐒𝐧⁑〖𝝅𝒙 𝒅𝒙〗 γ€— = [(βˆ’π‘₯ cos⁑〖(πœ‹π‘₯)γ€—)/πœ‹+sin⁑〖(πœ‹π‘₯)γ€—/πœ‹^2 ]_1^(3/2) = (((βˆ’3)/2 cos⁑(3πœ‹/2))/πœ‹+sin⁑((3πœ‹ )/2)/πœ‹^2 )βˆ’((βˆ’1 cosβ‘πœ‹)/πœ‹+sinβ‘πœ‹/πœ‹^2 ) = (0+((βˆ’1))/πœ‹^2 )βˆ’((βˆ’1 Γ— βˆ’1)/πœ‹+0/πœ‹^2 ) = (βˆ’1)/πœ‹^2 βˆ’1/πœ‹ Now, ∴ ∫_(βˆ’1)^(3/2)β–’|π‘₯ sin⁑(πœ‹π‘₯) | 𝑑π‘₯ = ∫_(βˆ’1)^1β–’γ€–π‘₯ sin⁑〖 (πœ‹π‘₯) 𝑑π‘₯ βˆ’βˆ«_1^(3/2)β–’γ€–π‘₯ 𝑠𝑖𝑛〗〗 γ€—(πœ‹π‘₯) 𝑑π‘₯ = 2/πœ‹βˆ’((βˆ’1)/πœ‹^2 βˆ’1/πœ‹) = 2/πœ‹+1/πœ‹^2 +1/πœ‹ = πŸ‘/𝝅+𝟏/𝝅^𝟐

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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.