How do you sketch the region enclosed by y=1+sqrtx, Y=(3+x)/3 and find the area?

1 Answer
Dec 26, 2016

4.5 square units

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 y = sqrt(x) + 1 is above y = (x + 3)/3. You can trace the square root function by drawing the function y = sqrt(x) and transform it one unit up. y = 1/3(x + 3) is equivalent to y = 1/3x + 1, which is your run of the mill linear function.

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