This gives the identityĭblquad demands a vectorized function (not symbolic expression) as an input, depending on vector-valued x and scalar-valued y. We set y=y1+u(y2-y1) so that as u varies from 0 to 1, y varies from y1 to y2, and then integrate over x and u instead of over x and y. make a change of variables in the inner integral.expand the region of integration to a rectangle, and make the integrand = 0 outside of the region where one wants to integrate,.There is a complication in using dblquad it does not accept variable limits. MATLAB has its own double integrator, called dblquad. The numerical evaluation in this case was performed using numerical routines borrowed from MuPAD, which takes care of symbolicĬalculations. However, we can also integrate any function we like over the same region, changing only The two functions of x whose graphs define the region that is to say the first integration in this case is so simple that one can write down the Of course, the area between two curves could just as easily have been evaluated as the integral of the difference between Since our limits for x are numerical, a symbolic calculation is not of much use directly, so we use double to convert to a numerical answer.Īrea=double(int(int(1,y1,y2),lims(1),lims(2))) To find the area of the region, for example, we integrate the function 1. We can now use them to integrate any function we like over the region in question. Title()Įvidently our limits define the right region. Gin=inline(vectorize(g+var-var),char(var)) % numbers and not just symbolic expressions like sqrt(2)įin=inline(,char(var)) %This guarantees these are floating point % in it, so if it's supposed to be a constant, try a fudge like % function g in blue, and lines are drawn connecting the two curves % of "Multivariable Calculus and Mathematica" for viewing the region %VERTICALREGION is a version for Matlab of the routine on page 157 Plots the two curves in two different colors: red for the function entered first, and blue for the function entered second,Īnd shades the region in between in cyan.įunction out = verticalRegion(a, b, f, g) Now that we have found the limits, we can use the following M-file verticalRegion to visualize the region they define so that we can check that it is the same as the region we had in mind. That seems to be the lower limit, but there should have been an upper limit around 1.7. Hmm., not clear what this means, but we can convert to numerical values: lims=double(ans) In order to find the limits for x, we need the values for which the two functions coincide. We can see from the plot that, to define the bounded region between the two graphs, exp(x) should be the lower limit for y, and 2x+2 should be the upper limit. Title( 'region bounded by y = 2x+2 and y = e^x') It should be large enough to show the region of interest, but small enough so that the region of interest occupies most of You may need to experiment with the interval to get a useful plot We begin by plotting the two curves on the same axes. For a first example, we will consider the bounded region between The limits we have chosen define the region we intended. While MATLAB cannot do that for us, it can provide some guidance through its graphics and can also confirm that We will now address the problem of determining limits for a double integral from a geometric description of the region of However, we can evaluate the integral numerically, using double.ĭouble(int(int(exp(x^2-y^2),y,1-x,1-x^2),x,0,1)) Warning: Explicit integral could not be found. In terms of erf(x), which is the (renormalized) antiderivative of exp(-x^2). However, if we change the integrand to, say,Įxp(x^2 - y^2), then MATLAB will be unable to evaluate the integral symbolically, although it can express the result of the first integration There is, of course, no need to evaluate such a simple integral numerically. We can even perform the two integrations in a single step: int(int(x*y,y,1-x,1-x^2),x,0,1) To evaluate the integral symbolically, we can proceed in two stages. We begin by discussing the evaluation of iterated integrals.Įxample 1 We evaluate the iterated integral
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