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  • Find the area of the region  <img alt="\left \{ \left ( x,y \right ):0\leq y\leq x^2, 0\leq y\leq x+2,-1\leq x\leq 3\right \}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%5C%7B%20%5Cleft%20%28%20x%2Cy%20%5Cright%20%29%3

Find the area of the region 

\left \{ \left ( x,y \right ):0\leq y\leq x^2, 0\leq y\leq x+2,-1\leq x\leq 3\right \}

 

 

 

 
 
 
 
 

Answers (1)

\left \{ \left ( x,y \right ):0\leq y\leq x^2, 0\leq y\leq x+2,-1\leq x\leq 3\right \}y=x^2,y=x+2,x=-1,x=3

Put y=0 in y=x+2

x+2=0

x=-2, x=0...> y=2

(-2,0)   (0,2)

Put y=x^2 in y=+2

x^2-x-2=0,x^2-2x+x-2=0

x-2+1(x-2)=0

x=2,-1

A=\int_{-1}^{2}\left ( x+2-x^2 \right )dx+\int_{2}^{3}\left ( x^2-x-2 \right )dx

= \left [ \frac{x^2}{2}+2x-\frac{x^3}{3} \right ]^2_{-1}+ \left [ \frac{x^3}{3}-\frac{x^2}{2}-2x \right ]^3_{2}

= \left [ \left ( \frac{18-8}{3} \right )-\left ( \frac{-7}{6} \right ) \right ]+\frac{11}{6}

= \left ( \frac{38}{6} \right )=\frac{19}{3}

Posted by

Ravindra Pindel

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