How do you sketch the region enclosed by y=1+sqrtx, Y=(3+x)/3 and find the area?
1 Answer
Explanation:
Start by finding the intersection points of the two curves.
1 + sqrt(x) = (3 + x)/3
3(1+ sqrt(x)) = 3 + x
3 + 3sqrt(x) =3 + x
3sqrt(x) = x+3 - 3
(3sqrtx)^2 = x^2
9x = x^2
0 = x^2-9x
0 = x(x-9)
x = 0 and 9
Next, sketch a rudimentary graph of the curves to see which one lies above the other. You will find the graph of
Integrate:
=int_0^9 sqrt(x) + 1 - (x + 3)/3 dx
=int_0^9 (3sqrtx + 3 - x - 3)/3dx
=int_0^9 (3sqrtx - x)/3 dx
=1/3int_0^9 3sqrtx -xdx
=1/3[2x^(3/2) - 1/2x^2]_0^9
=1/3(13.5)
=4.5 square units
Hopefully this helps