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  • Stuck here, help me understand: - Integral Calculus - JEE Main-3

Let g(x)=cos x2, f (x) = \sqrt{x} , and α, β (α < β) be the roots of the quadratic equation 18x2−9πx+π2=0. Then thearea (in sq. units) bounded by the curve y=(gof )(x) and the lines x=α, x=β and y=0, is :

  • Option 1)

    \frac{1}{2}\left ( \sqrt{2-1} \right )

  • Option 2)

    \frac{1}{2}\left ( {\sqrt3-1} \right )

  • Option 3)

    \frac{1}{2}\left ( {\sqrt3+1} \right )

  • Option 4)

    \frac{1}{2}\left ( {\sqrt3} -{\sqrt2} \right )

 

Answers (2)

best_answer

18x^{2}-9\pi x+\pi ^{2}=0

(3x-\pi ) (6x-\pi ) = 0 \Rightarrow x= \frac{\pi }{3}, x=\frac{\pi }{6}

thus \alpha =\frac{\pi }{6}, \beta = \frac{\pi }{3}

y= g\left ( f\left ( x \right ) \right ) = g\left ( \sqrt{x} \right )= \cos\left ( \sqrt{x} \right )^{2}= \cos x

 

Area = \int_{\pi /6}^{\pi /3} \cos x dx = \left ( \sin x \right)_{\pi /6}^{\pi /3}= \frac{\sqrt3}{2}- \frac{1}{2}

 

Geometrical integration of a definite integral -

An algebraic sum of the area of the figure bounded by the curve y = f(x), the x axis and the striaght lines x=a and x=b. The areas above x axis are taken as positive and the areas below x axis are taken as negative.  
 

- wherein

Where a< b

Hence

\int_{a}^{b}f\left ( x \right )dx=

ar\left ( PAQP \right )-ar\left ( QTRQ \right )+ar\left ( RBSR \right )

 

 


Option 1)

\frac{1}{2}\left ( \sqrt{2-1} \right )

Option 2)

\frac{1}{2}\left ( {\sqrt3-1} \right )

Option 3)

\frac{1}{2}\left ( {\sqrt3+1} \right )

Option 4)

\frac{1}{2}\left ( {\sqrt3} -{\sqrt2} \right )

Posted by

Himanshu

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