Verified answer

I presume the "element of the real numbers squared" sign means both x and y are an element of the real numbers - I do physics, not maths!

I got 7/24 for the first question with the same double integral as you have. If I'm correct, I presume you lost a minus somewhere or did the integration in the wrong order. My graph was plotted as x = y^2 for the bounding curve.

For the 2nd question I got 1/4 with a different set of limits for the dx integral (which I did first): 0 to sqrt(y). My graph was the same as yours.

I always think of integrating over strips of width dx and dy in which ever order is most convenient, until the entire area is covered.

From the additional information you will have given, it just appears like you're going the correct way except that you've put the x and x^2 limits the incorrect approach round. Here is what you must have (roughly): So we're integrating with admire to (w.R.T) y first after which x. That implies our inner limits could also be features of x, however the outer limits have to be consistent values. The imperative we are actually doing is: intfrom x = zero to 1 [ intfrom y = x^2 to y = x [x + y]dy]dx. (in case you get the notation.) Doing the interior necessary first: (x + y) integrates by y to provide (half of)y^2 + xy. Taken between the bounds y = x (high limit) and y = x^2 (backside limit), that offers (half of)x^2 + x*x - (half of)(x^2)^2 + x*x^2 = (3/2)x^2 - (half)x^four + x^3. NB! The limit of x is put in first, and then the limit of x^2, considering over the variety [0, 1], x >= x^2. In the event you do them the wrong way circular you get the poor of the correct answer! Now we combine this w.R.T x: -> (half)x^three - (1/10)x^5 + (1/four)x^4. Now putting in the limits of 1 and zero, we get -> [(1/2)*1^3 - (1/10)*1^5 + (1/4)*1^4] - [(1/2)*0^3 - (1/10)*0 + (1/4)*0^4] = half of - 1/10 + 1/4 - [0 - 0 + 0] = thirteen/20.