Question:

The area of the region in the first quadrant enclosed by the curvesy=x,y=x+6 y=√x,y=-x+6 and the x-axis is 

Updated On: Dec 10, 2024
  • 227\dfrac{22}{7}

  • 223\dfrac{22}{3}

  • 1212

  • 2424

  • 88

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The Correct Option is B

Solution and Explanation

Given that

y=x,y = √ x,    y=x+6y = −x + 6

Then

x=3612x+x2x = 36 − 12x + x^2

x213x+36=0x^2 − 13x + 36 = 02

(x4)(x9)=0(x-4)(x-9)=0

 x=4\therefore x=4 and x=9 x=9

represents area of the region in the first quadrant enclosed by the curves y=√x, y = −x+6 and the x-axis

Then, Area=06f(x)dx = ∫_0^6 f(x)dx

=06f(x)dx= ∫_0^6 f(x)dx

=04xdx+46(x+6)dx=∫_0^4√xdx+∫_4^6(-x+6)dx

=[23x32]04+[x22+6x]46=[\dfrac{2}{3}x^{\dfrac{3}{2}}]_0^4+[\dfrac{-x^2}{2}+6x]_4^6

=163+18+824=\dfrac{16}{3} + 18 + 8 − 24

=223= \dfrac{22}{3} (_Ans)

 

 

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Area under Simple Curves

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