Log in to add a comment

## Answers

to get the answer first find the intersecting points(3,0) of both curves then integrate the both equations then put the limits 0 to 3 and solve it

→x² −x −2 =0 i.e (x−2)(x+1)=0

I.e either x=2 or x= −1

hence Area of the region bounded by y=2x−x^2 and y=x−2

is A=∫ (from x= −1_to x=2) of {(2x−x^2) −(x−2)}dx

A=∫ (from x= −1_to x=2) of {(x−x^2+2) }dx

A= {(x²−x³/3+2x)} (from x= −1_to x=2)

= {(2²−2³/3+2×2)} − {(−1)²−(−1)³/3+2(−1)}

= {(4−8/3+4)} − {1+1/3−2)}

= {(16/3)} +{2/3} = 6 sq. units