Hi I'm new and a calculus moron. Can someone explain how to double integrate this step by step? Thank you
Evaluate the double integral over the given region R.
f(x, y) = 6x^(3x^2+y); R is the rectangle defined by 0< x < 1 and -3 < y < 0
Hi I'm new and a calculus moron. Can someone explain how to double integrate this step by step? Thank you
Evaluate the double integral over the given region R.
f(x, y) = 6x^(3x^2+y); R is the rectangle defined by 0< x < 1 and -3 < y < 0
I won't explain this step by step, but I'll post something potentially useful until someone else helps you out more.
So with double integrals the difficulties can be with setting up the limits to describe the region of integration, R, and choosing the order in which you integrate. Now for this problem you are given the limits which is very nice. It is also good because you are integrating over a rectangle with fixed endpoints. Sometimes you have to integrate over regions defined by other functions, like y=x, and this becomes more tricky.
Since you have your region and limits already given to you, now you must decide in what order to integrate. Which order is best? dxdy or dydx?
I'm not sure because I don't know how to integrate the function itself, regardless of what variables I'm plugging in later. If I use dxdy, when I plug in the limits for x, I end up with something that I don't know how to integrate again. If I use dydx, I don't know how to integrate it when the variable is in the exponent
I mean : are you sure that you need to integrate $\displaystyle 6x^{3x^2+y}$
One part is feasible
$\displaystyle \int_0^1\int_{-3}^0 6x^{3x^2+y}\:dy\: dx = \int_0^1 \int_{-3}^0 6e^{(3x^2+y)\:\ln x}\:dy \:dx = \int_0^1 \left[\frac{6e^{(3x^2+y)\:\ln x}}{\ln x}\right]_{y=-3}^{y=0} \:dx$