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Problem 59

Recall from the Volume and Average Value section …

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Problem 58

Evaluate each double integral. If the function seems too difficult to integrate, try interchanging the limits of integration, as in Exercises 37 and $38 .$
$$\int_{0}^{2} \int_{y / 2}^{1} e^{x^{2}} d x d y$$

Answer

$e-1$



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Video Transcript

Okay, so we have a double in a girl zero to and why, over too toe one of either the X squared the ex. See why this is a classic example of if you tried Teo, do the integral first with respect to X. This is impossible s in the X squared does not have an elementary. The anti drug It's, I mean, you could write out, like Taylor Siri's or something and doing approximation. But, I mean, you can not just find a simple function. You know, Nina, whatever. Who's anti derivatives? Who's driven is going to be eating the X squared. So we need to switch to why first and hope that something good happens. So let's actually draw the region because we're going to switch the limits or switch X and Y. We need to re express our limits of integration, sir. Right now we have some why fixed between zero and two. Can I see yours? Zira two. And here's their wide value. And why's is supposed to be going from the function? Why over to so X equals y over too, which is the same as the function y equals two X that's going to be here at two. X equals one in Mexico like that. Okay. And so why is going from Why over too? You No one region here. So if we want to first fix an X value, we're going to be fixing the next value between zero and one and going from zero up to this line. Why? He calls to X for Michael zero y equals two x So every sledge we'LL have X going from zero to one on then for a fixed X Y is going from zero to X when they have each of the X squared. Dee, Why the ex? Now, look, it was gonna happen here. This in the X squared is constant. This is amazing. So you're either the X squared, and then we're just integrating d Y from zero to two. Ex. Well, that's just going to give us a factor of two X. But now, look this We know what the anti driven out of this is that this works out perfectly. So the derivative of E to the X squared is e to the X squared times to X. So this equals into the X squared. Just evaluated from zero one. That's just yeah, minus one

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