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.