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Example 43 - Evaluate integral |x sin (pi x)| dx - Limit -1 to 3pi/2

Example 43 - Chapter 7 Class 12 Integrals - Part 2
Example 43 - Chapter 7 Class 12 Integrals - Part 3 Example 43 - Chapter 7 Class 12 Integrals - Part 4 Example 43 - Chapter 7 Class 12 Integrals - Part 5 Example 43 - Chapter 7 Class 12 Integrals - Part 6 Example 43 - Chapter 7 Class 12 Integrals - Part 7 Example 43 - Chapter 7 Class 12 Integrals - Part 8 Example 43 - Chapter 7 Class 12 Integrals - Part 9 Example 43 - Chapter 7 Class 12 Integrals - Part 10

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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‘₯ ≀ [email protected]&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‘₯ ≀ [email protected]&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 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.