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  • Prove that the curves and divide the area of square bounded

Prove that the curves y^{2}= 4x and x^{2}= 4y divide the area of square bounded by x= 0,x= 4,y= 4 and y= 0 into three equal parts.

 

 

 

 
 
 
 
 

Answers (1)

Let A1,A2 & A3 denote areas OSPQO, OSPTO and OTPRO resp. To prove: A_{1}= A_{2}= A_{3}
Now, A_{1}= \int_{0}^{4}\frac{x^{2}}{4}dx= \frac{1}{4}\int_{0}^{4}x^{2}dx
= \frac{1}{4}\left [ \frac{x^{3}}{3} \right ]^{4}_{0}= \frac{1}{4}\times \frac{64}{3}= \frac{16}{3}\, sq.units

A_{2}=\left ( Area\; bounded\; by\; y^{2}= 4x \right )-\left (Area\; bounded\; by\; x^{2}= 4y \right )
A_{2}=\int_{0}^{4}\left ( \sqrt{4x}-\frac{x^{2}}{4} \right )dx
=\int_{0}^{4}\left ( 2\sqrt{x}-\frac{x^{2}}{4} \right )dx\Rightarrow \left [ \frac{4}{3}x^{\frac{3}{2}}-\frac{x^{3}}{12} \right ]^{4}_{0}
=\left [ \frac{4}{3}\times 8-\frac{64}{12} \right ]= \frac{16}{3}sq\cdot units
A_{3}=\int_{0}^{4}\frac{y^{2}}{4}dy
= \frac{1}{4}\left [ \frac{y^{3}}{3} \right ]^{4}_{0}= \frac{1}{4}\times \frac{64}{3}= \frac{16}{3}sq.units
Hence A1 =A2=A3 Hence proved

Posted by

Ravindra Pindel

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