Area as a sum

Chapter 7 Class 12 Integrals
Serial order wise

This Question was also asked in CBSE Maths Board Exam - 2020 (Question 34 - Set 65/5/1)

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### Transcript

Question 4 β«1_1^4β(π₯2 βπ₯)ππ₯ Let I = β«1_1^4β(π₯2 βπ₯)ππ₯ I = β«1_1^4βγ π₯2 ππ₯γββ«1_1^4βγ π₯ ππ₯γ Solving I1 and I2 separately Solving I1 β«1_1^4βγπ₯2 ππ₯γ Putting π =1 π =4 β=(π β π)/π =(4 β 1)/π =3/π π(π₯)=π₯^2 We know that β«1_π^πβγπ₯ ππ₯γ =(πβπ) (πππ)β¬(πββ) 1/π (π(π)+π(π+β)+π(π+2β)β¦+π(π+(πβ1)β)) Hence we can write β«1_1^4βγπ₯2 ππ₯γ =(4β1) (πππ)β¬(πββ) 1/π (π(1)+π(1+β)+π(1+2β)+ β¦+π(1+(πβ1)β)) =3 (πππ)β¬(πββ) 1/π (π(1)+π(1+β)+π(1+2β)+ β¦+π(1+(πβ1)β)) Here, π(π₯)=π₯^2 π(1)=(1)^2=1 π(1+β)=(1+β)^2 π (1+2β)=(1+2β)^2 β¦ π(1+(πβ1)β)=(1+(πβ1)β)^2 Hence, our equation becomes β«1_1^4βγπ₯2 ππ₯γ " " =3 (πππ)β¬(πββ) 1/π (π(1)+π(1+β)+π(1+2β)+ β¦+π(1+(πβ1)β)) =3 (πππ)β¬(πββ) 1/π ((1)^2+(1+β)^2+(1+2β)^2+ β¦+(1+(πβ1)β)^2 ) =3 (πππ)β¬(πββ) 1/π (β(1^2+(1^2+β^2+2β)+γ(1γ^2+ (2β)^2+4β)+ β¦β¦ @ β¦+(1^2+((πβ1)β)^2+2(πβ1) β) )) =3 (πππ)β¬(πββ) 1/π [1^2+1^2+ β¦ +1^2 ] + β^2+(2β)^2+ β¦ +(πβ1)β^2 + [2β+4β+ β¦ +2(πβ1)β] =3 (πππ)β¬(πββ) 1/π (γπ(1)γ^2+[β^2+(2)^2 . β^2+ β¦ +(πβ1)^2 β^2 ] +[2β+2Γ2β+ β¦ +(πβ1)Γ2β] ) =3 (πππ)β¬(πββ) 1/π(π+π^2 [(1)^2+(2)^2+ β¦+(πβ1)^2 ] +ππ [1+2+ β¦+(πβ1)]) =3 (πππ)β¬(πββ) 1/π (π+β^2 [π(π β 1)(2π β 1)/6]+2β[π(π β 1)/2] ) We know that 1^2+2^2+ β¦+π^2= (π (π + 1)(2π + 1))/6 1^2+2^2+ β¦β¦+(πβ1)^2 = ((π β 1) (π β1 + 1)(2(π β 1) + 1))/6 = ((π β 1) π (2π β 2 + 1) )/6 = (π (π β 1) (2π β 1) )/6 We know that 1+2+3+ β¦β¦+π= (π (π + 1))/2 1+2+3+ β¦β¦+(πβ1) = ((π β 1) (π β 1 + 1))/2 = (π (π β 1) )/2 =3 (πππ)β¬(πββ) 1/π (π+β^2 [π(π β 1)(2π β 1)]/6+β[π(π β 1)] ) =3 (πππ)β¬(πββ) (π/π+β^2 [π(π β 1)(2π β 1)/6π]+β[π(π β 1)/π]) =3 (πππ)β¬(πββ) (1+β^2 [(π β 1)(2π β 1)/6]+β[(π β 1)]) =3 (πππ)β¬(πββ) (1+(3/π)^2 (π β 1)(2π β 1)/6+(3/π)(π β 1)) =3 (πππ)β¬(πββ) (1+9/π^2 . (π β 1)(2π β 1)/6 +3(1 β 1/π)) =3 (πππ)β¬(πββ) (1+ 9(1 β 1/π)(2 β 1/π)/6 +3(1 β 1/π)) =3(1+ 9(1 β 1/β)(2 β 1/β)/6 +3(1 β 1/β)) =3(1+ 9(1 β 0)(2 β 0)/6 +3(1 β0)) =3(1+ (9 Γ 1 Γ 2)/6 +3) =3(1+3+3) =3Γ7 =ππ Solving I2 β«1_1^4βγπ₯ ππ₯γ Putting π =1 π =4 β=(π β π)/π =(4 β 1)/π =3/π π(π₯)=π₯ We know that β«1_π^πβγπ₯ ππ₯γ =(πβπ) (πππ)β¬(πββ) 1/π (π(π)+π(π+β)+π(π+2β)β¦+π(π+(πβ1)β)) Hence we can write β«1_1^4βγπ₯ ππ₯γ =(4β1) limβ¬(nββ) 1/π (π(1)+π(1+β)+π(1+2β)+β¦ +π(1+(πβ1)β) =3 limβ¬(nββ) 1/π (π(1)+π(1+β)+π(1+2β)+β¦ +π(1+(πβ1)β) Here, π(π₯)=π₯ π(1)=1 π(1+β)=1+β π (1+2β)=1+2β π(1+(πβ1)β)=1+(πβ1)β Hence, our equation becomes β«_1^4βπ₯ ππ₯ =3 limβ¬(nββ) 1/π (π(1)+π(1+β)+π(1+2β)+β¦ +π(1+(πβ1)β) = 3 (πππ)β¬(πββ) 1/π (1+(1+β)+(1+2β)+ β¦+(1+(πβ1)β)) = 3 (πππ)β¬(πββ) 1/π (1+1+ β¦+1 +β+2β+ β¦β¦+(πβ1)β) = 3 (πππ)β¬(πββ) 1/π ( π\ Γ1+β (1+2+ β¦β¦β¦+(πβ1))) We know that 1+2+3+ β¦β¦+π= (π (π + 1))/2 1+2+3+ β¦β¦+πβ1= ((π β 1) (π β 1 + 1))/2 = (π (π β 1) )/2 = 3 (πππ)β¬(πββ) 1/π ( π+(β . π(π β 1))/2) = 3 (πππ)β¬(πββ) ( π/π+π(π β 1)β/2π) = 3 (πππ)β¬(πββ) ( 1+(π β 1)β/2) = 3 (πππ)β¬(πββ) ( 1+(π β 1)3/(2 . π)) = 3 (πππ)β¬(πββ) ( 1+(π/π β 1/π) 3/2) [ππ πππ β=3/π] = 3 (πππ)β¬(πββ) ( 1+(1β 1/π) (3 )/2) = 3( 1+(1β 1/β) (3 )/2) = 3( 1+(1β0) 3/2) = 3(1+ (3 )/2) = 3((5 )/2) = ππ/π Putting the values of I1 and I2 in I β΄ "I = " β«1_1^4βγ π₯2 ππ₯γββ«1_1^4βγ π₯ ππ₯γ = 21 β 15/2 = (42 β 15)/2 = ππ/π