# Thread: 2 Area between 2 curves problems

1. ## 2 Area between 2 curves problems

There are two problems that no matter what I attempt to do, can not acquire the right answer for on my current assignment , I hope that someone here can quickly help me figure them out.

The total area enclosed by the graphs of y=12x^2 – x^3 + x
y=x^2 + 31x

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

2y+2x=5, y = 3, 2y = 3(x)^(1/2)

2. Surprised nobody's gotten to this yet.

For the first one:

The two curves intersect at x=0 and x=1.048, according to the graphs on my calculator and its intersect-finder. This means that, integrating with respect to x, the lower limit of integration should be 0 and the upper limit should be 1.048. The values of the second function, y=x^2 + 31x, are higher over this interval, so the area can be found by taking the integral (with the previously described limits of integration) of the second function minus the first function, giving us "integral from 0 to 1.048 of (x^2 + 31x - (12x^22 – x^3 + x))dx = integral from 0 to 1.048 of (-12x^22 + x^3 + x^2 + 30x). I will define the integrand as g(x) and say that the integral of it is, note the capital G, G(x) = (-12/23)x^23 + (1/4)x^4 + (1/3)x^3 + 15x^2. Evaluated from the lower limit to the upper, that yields G(1.048) - G(0) = G(1.048) = 15.626.

Did you get all that

Now for the second problem:

2y+2x=5, y = 3, 2y = 3(x)^(1/2)

To make things more clear, let's put the first and last equations in explicit form (that is, isolate y.)

The first equation becomes y = (1/2)(5 - 2x) = -x + 5/2, a line with slope -1 and y-intercept 5/2. The third equation becomes y = (3/2)x^(1/2) = (3/2)sqrt(x). Now find the point(s) of intersection by setting them equal to each other: -x + 5/2 = (3/2)sqrt(x). Solve for x and you get 1. So that leaves us with a somewhat triangular region above both curves and below the line y = 3. Integrating with respect to y will require only one integral (whereas with respect to x would require two since the borders of the region are two different functions, switching at the point of intersection.) So now isolate x in both equations. The first becomes x = -y + 5/2 and the second becomes x = (4/9)y^2. The second equation is "higher;" that is, its x value is more positive. Therefore, the area between the curves can be found by integrating the second equation minus the first equation, from 3/2 (the y-coordinate of the intersect point) to y=3 (the upper border of the region in question.) Integral from 3/2 to 3 of (4/9)y^2 - (-y + 5/2) = Integral from 3/2 to 3 of (4/9)y^2 + y - 5/2. If the integrand is g(y) then the resulting function is G(y) = (4/27)y^3 + (1/2)y^2 - (5/2)y. The value of the definite integral is given by G(3) - G(3/2), which you'll have to evaluate yourself because I need to go to bed, like, now. Hope that helps.

3. Oh no! I'm very sorry, but there was a really bad typo in my original post, it should be 12x^2 instead of 12x^22, really bad mistake on my part there.

4. Lol, I thought something like that might be the case. I thought that was a rather strange problem to be assigned. To save the trouble of retyping it, I'm just going to adjust the numbers and copy and paste.

The two curves intersect at x=0 and x=6, according to the graphs on my calculator and its intersect-finder. This means that, integrating with respect to x, the lower limit of integration should be 0 and the upper limit should be 6. The values of the second function, y=x^2 + 31x, are higher over this interval, so the area can be found by taking the integral (with the previously described limits of integration) of the second function minus the first function, giving us "integral from 0 to 6 of (x^2 + 31x - (12x^2 – x^3 + x))dx = integral from 0 to 1.048 of (x^3 - 11x^2 + 30x). I will define the integrand as g(x) and say that the integral of it is, note the capital G, G(x) = (1/4)x^4 - (11/3)x^3 + 15x^2. Evaluated from the lower limit to the upper, that yields G(6) - G(0) = G(6) = 72.

Now that's more the kind of nice clean answer I'd expect from a school problem.