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  • Using integration, find the area of the region <img alt="\left \{ \left ( x,y \right ):0\leq 2y\leq x^{2},0\leq y\leq x,0\leq x\leq 3 \right \}\cdot" src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%5C%7B%20%5Cleft%20%28%20x%2Cy%20%5Cright

Using integration, find the area of the region \left \{ \left ( x,y \right ):0\leq 2y\leq x^{2},0\leq y\leq x,0\leq x\leq 3 \right \}\cdot

 

 

 

 
 
 
 
 

Answers (1)

consider the diagram below:

consider the coresponding equations:
2y= x^{2}---(i)\,\, y= x---(ii)
x= 3---(iii)\; and\; y= 0,x= 0
on solving (i) and (ii) we get:
(0,0) , (2,0)
Also, on solving (i) and (iii) we get: \left ( 3,\frac{9}{2} \right )\cdot
And the point of intersection of (ii) and (iii) will be at (3,3)
Now the required area = \int_{0}^{2}\frac{x^{2}}{2}dx+\int_{2}^{3}x\, dx= \frac{1}{6}\left [ x^{3} \right ]^{2}_{0}+\frac{1}{2}\left [ x^{2} \right ]^{3}_{2}
              = \frac{1}{6}\left [ 8-0 \right ]+\frac{1}{2}\left [ 9-4 \right ]= \frac{4}{3}+\frac{5}{2}= \frac{23}{6}\, sq\, units                 

Posted by

Ravindra Pindel

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