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Sketch the region between V =2 and the X-axis over the interval [ ~ 3 , 3]...

Question

Sketch the region between V =2 and the X-axis over the interval [ ~ 3 , 3]

Sketch the region between V = 2 and the X-axis over the interval [ ~ 3 , 3]



Answers

Sketch the region described and find its area. The region below the interval $[-2,-1]$ and above the curve $y=x^{3} .$

No basis for this problem. They want you to integrate the function. Weichel soup, extra pulse one along the interval from 03 Andi also want you to sketch that area over here. I use Desmond's to pot without function. Why? Execs were plus one. And we're integrating from X equals zero to X equals 23 So our area would be this region right here, all the way up. And it keeps going all the way up on toe. Uh, 10 were y equals 2 10 So if you want to integrate this function, we have to All right. In black, you have integral from 0 to 3 of x squared plus one the X. And if we were to find the anti derivative of listen to grill, it would be excused over three plus x from the bouncer from 0 to 3. So if we were to plug in our boundaries for our boundary values for X, we get three cubed over three, which is 27 or three, which is nine plus three. When people so 12. Because no, I billion through we just give you zero. So this is your final answer.

X between minus 2 to 3, minus three to plus tree. Let's say we'll have Why exits we have XX is it is autism. 123 minus one minus two minus street and we'll draw the line minus three and X equals two plus three. All the reason between this point that is these two lines are represents this area or this reason which inclusive and which tells the X values always between minus treated, say so it is open from why from minus and review, plus infinity but X is wounded. But in my industry, two x equals two plus off street.

For this question were given the following soured absolute rally of boxes lost or England to an absolute value wise less than equal to three. So we think about to draw their lines. Then we have severe abso value three. When we knows that both of them are luster equal to you. So to fit both of our parameters are graph looks something like this. So this is our answer.

This question were given The phone Lee set when I get you is less an X, which is awesome too. So we can search for drawing out that. So we have. So he seems less than she does these air dollars lines. We also snow that y squared equals three. So we know that our that is part of this region. My graph. So this is our answer.


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