Misc 6 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 12, 2019 by
Last updated at Dec. 12, 2019 by
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Misc 6 Find the area enclosed between the parabola π¦2=4ππ₯ and the line π¦=ππ₯ Letβs first draw the Figure Here, π¦2 =4ax is a Parabola And, π¦=ππ₯ is a straight line Let A be point of intersection of line and parabola Finding point A Putting y = mx in equation of parabola π¦^2=4ππ₯ (ππ₯)^2=4ππ₯ π^2 π₯^2=4ππ₯ π^2 π₯^2β4ππ₯=0 π₯(π^2 π₯β4π)=0 Therefore, π₯=0 π^2 π₯β4π=0 π^2 π₯=4π π₯=4π/π^2 Putting values of π₯ in π¦=ππ₯ π¦=π Γ0=0 π¦=π Γ4π/π^2 =4π/π So, the intersecting points are O(π , π) and A (ππ/π^π ,ππ/π) Finding Area Area Required = Area OBAD β Area OAD Area OBAD Area OBAD = β«_0^(4π/π^2 )βγπ¦ ππ₯" " γ y β Equation of parabola π¦^2 = 4ax π¦ = Β± β("4" ππ₯) Since OBAD is in 1st quadrant, value of y is positive β΄ y = β("4" ππ₯) Now, Area OBAD =β«_0^(4π/π^2 )βγβ4ππ₯ ππ₯" " γ =β«_0^(4π/π^2 )βγβ4π .βπ₯ ππ₯" " γ =β4π β«_0^(4π/π^2 )βγβπ₯ ππ₯" " γ =2βπ [π₯^(1/2 + 1)/(1/2 + 1)]_0^(4π/π^2 ) =β4π [π₯^(3/2)/(3/2)]_0^(4π/π^2 ) =β4π Γ 2/3 [π₯^(3/2) ]_0^(4π/π^2 ) =(2(2βπ))/3 [(4π/π^2 )^(3/2)β(0)^(3/2) ] =(4βπ)/3 [4π/π^2 β(4π/π^2 )β0] =(4βπ)/3 [4π/π^2 Γ(2βπ)/π] =(32 π .βπ .βπ)/(3π^3 ) =(32 π^2)/(3π^3 ) Area OAD Area OAD = β«1_0^(4π/π^2 )βγπ¦ ππ₯γ y β Equation of line y = mx Therefore, Area OAD = β«1_0^(4π/π^2 )βγ"m" π₯ ππ₯γ = mβ«1_0^(4π/π^2 )βγπ₯ ππ₯γ = π[π₯^2/2]_0^(4π/π^2 ) = π[π₯^2/2]_0^(4π/π^2 ) =π/2 [(4π/π^2 )^2β0^2 ] =π/2 (4π)^2/π^4 =(8π^2)/π^3 Thus, Area Required = Area OBAD β Area OAD = (32 π^2)/(3π^3 ) β (8π^2)/π^3 = ((32 β 24) π^2)/(3π^3 ) = (ππ^π)/γππγ^π
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