Question

area enclosed by y=x^2 and y=|x|+2​

Answer 1

Given :

The equation is  y = x²   and    y = $\left | x \right |$ + 2

To Find :

The Area enclosed by y = x²   and    y = $\left | x \right |$ + 2

Solution :

  y = x²              ..........1

And  y = x + 2              ..........2

And  y = - x + 2              .......3

Solving eq 1 and eq 2

x² = x + 2

Or,  x² - x - 2 = 0

Or,  x² - 2 x + x - 2 = 0

or,  x ( x - 2 ) + 1 ( x - 2 ) = 0

∴  ( x - 2 )  ( x + 1 ) = 0

i.e   x = 2 , - 1

Again

Solving eq 1 and eq 3

x² = - x + 2

Or,  x² + x - 2 = 0

Or,  x²  2 x - x - 2 = 0

or,  x ( x + 2 ) - 1 ( x + 2 ) = 0

∴  ( x + 2 )  ( x - 1 ) = 0

i.e   x = - 2 ,  1

Area enclosed = $\int_{-1}^{2}( x^{2}-x-2)dx$ + $\int_{-2}^{1}( x^{2}+x-2)dx$

                        = [ $\dfrac{x^{3} }{3}$ - $\dfrac{x^{2} }{2}$ - 2 x ] ( - 1 to 2 ) + [ $\dfrac{x^{3} }{3}$ + $\dfrac{x^{2} }{2}$ - 2 x ] ( -2 to 1 )

                        = $\dfrac{(2)^{3}-(-1)^{3} }{3}$ - $\dfrac{(2)^{2}-(-1)^{2} }{2}$  - 2 (  2 + 1) + $\dfrac{(-2)^{3}-(1)^{3} }{3}$ -

                         $\dfrac{(-2)^{2}-(1)^{2} }{2}$  - 2 (  -2  + 1)

                       = ( $\dfrac{9}{3}$ ) - ( $\dfrac{3}{2}$ ) - 2 ( 3 ) + ( $\dfrac{-9}{3}$ ) -  ( $\dfrac{-5}{2}$ ) - 2( - 1)

                       = ( 0 ) + ( $\dfrac{-3+5}{2}$ ) - 6 + 2

                       =  ( $\dfrac{2}{2}$ ) - 6+2

                       = 1 - 6 +2

                       = - 3

So, Area enclosed = $\left | -3|\right$ = 3 sq unit

Hence, The Area enclosed by given equation is 3 sq unit   Answer