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Question 14 of 19

Q.Find the area between the two curves (parabolas)
y2=7xy^2 = 7x and x2=7yx^2 = 7y.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★
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Find the intersection points (0,0)(0,0) and (7,7)(7,7), then integrate (7x−x27)\big(\sqrt{7x}-\tfrac{x^2}{7}\big) from 00 to 77 to get 493\dfrac{49}{3} sq. units.

Points of intersection. From y2=7xy^2=7x we get y=7xy=\sqrt{7x}, and from x2=7yx^2=7y we get y=x27y=\dfrac{x^2}{7}. Equating:

7x=x27 ⇒ 49⋅7x=x4 ⇒ x4=343x ⇒ x(x3−343)=0\sqrt{7x}=\dfrac{x^2}{7}\ \Rightarrow\ 49\cdot 7x = x^4\ \Rightarrow\ x^4=343x\ \Rightarrow\ x(x^3-343)=0,

so x=0x=0 or x=7x=7. The curves meet at (0,0)(0,0) and (7,7)(7,7).

Which curve is on top. For 0<x<70<x<7, the parabola y=7xy=\sqrt{7x} lies above y=x27y=\dfrac{x^2}{7} (e.g. at x=1x=1, 7≈2.65>17\sqrt7\approx 2.65 > \tfrac17).

Set up and evaluate the area.

A=∫07(7x−x27)dx=∫077 x1/2 dx−∫07x27 dxA=\displaystyle\int_0^7\left(\sqrt{7x}-\dfrac{x^2}{7}\right)dx=\int_0^7\sqrt7\,x^{1/2}\,dx-\int_0^7\dfrac{x^2}{7}\,dx.

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