Area under curve

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Ex 8.1, 5 Find the area of the region bounded by the ellipse π₯^2/4+π¦^2/9=1 Given Equation of Ellipse π₯^2/4+π¦^2/9=1 π₯^2/(2)^2 +π¦^2/(3)^2 =1 Area of Ellipse = Area of ABCD = 2 Γ [Area of BCD] = 2 Γ β«_(β2)^2βγπ¦ ππ₯γ We know that π₯^2/4+π¦^2/9=1 π¦^2/9=1βπ₯^2/4 π¦^2/9=(4 β π₯^2)/4 π¦^2=9/4 (4βπ₯^2 ) Taking Square Root on Both Sides π¦="Β±" β(9/4 (4βπ₯^2 ) ) π¦ ="Β±" 3/2 β(4βπ₯^2 ) Since Area BCD is above the x-axis, The value of y will be positive β΄ π¦=3/2 β(4βπ₯^2 ) Now, Area of Ellipse = 2 Γ β«_(β2)^2βγπ¦ ππ₯γ = 3 β«_(β2)^2βγ β((2)^2βπ₯^2 ) ππ₯γ = 3 [π₯/2 β((2)^2βπ₯^2 )+(2)^2/2 sin^(β1)β‘γπ₯/2γ ]_(β2)^2 It is of form β(π^2βπ₯^2 ) ππ₯=π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Replacing a by 2 we get = 3 [π₯/2 β(4βπ₯^2 )+2 sin^(β1)β‘γπ₯/2γ ]_(β2)^2 = 3[(2/2 β(4β2^2 )+2 sin^(β1)β‘γ2/2γ )β((β2)/2 β(4βγ(β2)γ^2 )+2 sin^(β1)β‘γ(β2)/2γ )] = 3[(0+2 sin^(β1)β‘1 )β(0+2 sin^(β1)β‘γ(β1)γ )] = 3[2 sin^(β1)β‘1β2 γπππγ^(βπ)β‘γ(βπ)γ ] = 3[2 sin^(β1)β‘1β2Γβγπππγ^(βπ)β‘π ] = 3[2 γπ ππγ^(β1)β‘1+2 γπ ππγ^(β1)β‘1 ] = 3 [4 γπππγ^(βπ)β‘(π) ] = 3 Γ 4 Γ π/π = 6Ο β΄ Area of Ellipse = 6Ο square units (As sin-1 (βΞΈ) = βsinβ1 ΞΈ )