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The area (in sq. units) of the region described by

\left \{ (x,y):y^{2}\leq 2x\, and\, y\geq 4x-1 \right \}\, \; is:

  • Option 1)

    \frac{7}{32}

  • Option 2)

    \frac{5}{64}

  • Option 3)

    \frac{15}{64}

  • Option 4)

    \frac{9}{32}

 

Answers (1)

As learnt in concept

Area along x axis -

Let y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x) be two curve then area bounded between the curves and the lines

x = a and x = b is

\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |

 

- wherein

Where \Delta y= f_{2}\left ( x \right )-f_{1}(x)

 

 Ponits of intersection of y^2=2x \: and \: y=4x-1

2x= (4x-1)^2\Rightarrow 2x=16x^2-8x+1

\Rightarrow 16x^2 - 10x +1=0 \Rightarrow 16x^2 - 8x -2x+1=0

x=\frac{1}{2}, \frac{1}{8}

\int_{\frac{1}{8}}^{\frac{1}{2}}\left (\sqrt{2x}-(4x-1) \right )dx

\frac{\sqrt{3}}{\frac{3}{2}}(x^\frac{3}{2})^\frac{1}{2}_\frac{1}{8} - 2(x^2)^\frac{1}{2}_\frac{1}{8} -(x)^\frac{1}{2}_\frac{1}{8}

=\frac{9}{32}


Option 1)

\frac{7}{32}

Incorrect

Option 2)

\frac{5}{64}

Incorrect

Option 3)

\frac{15}{64}

Incorrect

Option 4)

\frac{9}{32}

Correct

Posted by

Sabhrant Ambastha

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