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jocelynevsq
Group Title
idenify the definite integral that represents the area of the region bounded by the graphs of y=x and y=5xx^3
 one year ago
 one year ago
jocelynevsq Group Title
idenify the definite integral that represents the area of the region bounded by the graphs of y=x and y=5xx^3
 one year ago
 one year ago

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cherio12 Group TitleBest ResponseYou've already chosen the best response.0
try graphing them first to determine which function is the top and which is the bottom
 one year ago

jocelynevsq Group TitleBest ResponseYou've already chosen the best response.0
still don't get it
 one year ago

jocelynevsq Group TitleBest ResponseYou've already chosen the best response.0
dw:1359920225466:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So ummmm yah that graph looks accurate. I always get confused by these types of problems. When the upper and lower functions switch places. But I think we can figure it out. Hmm.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Let's start by finding points of intersection.\[\large y=x \qquad y=5xx^3\qquad \qquad \rightarrow \qquad \qquad x=5xx^3\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Which turn out to be, \(\large x=0, \quad x=2, \quad x=2\).
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Let's look at just the right side for a moment, maybe it will make a little more sense that way.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1359921446750:dwSee how I've drawn a small rectangle? Let's find the area of that rectangle.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
The area will be `base x height`. The base has a thickness of \(dx\). The height is going to be `the upper function` minus `the lower function`.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large A=\left(height\right)(base)\]\[\large A=\left(5xx^3x\right)(dx)\]See how the line y=x is BELOW the other one? So we subtract y=x from the other function to get the correct height. So this represents the area of a small rectangular slice. Now we want to Integrate (sum up) all of the rectangles between these two curves.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
And we'll include our intersecting points.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \int\limits_0^2 4xx^3 \; dx\] This represents the area between the two curves on the RIGHT side. We still need to find the area of the other chunk. Notice how they switch places on the left? The line y=x is now the UPPER function! So we would subtract y=5xx^3 from the line y=x. And our limits of integration would be a little different.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
There is a little trick we could pull though c: See how they're both odd functions? They're both symmetric about the origin. So the area on the left will be the same as the area on the right. We could simply multiply this integral by 2.\[\huge A_{total}\qquad =\qquad 2\int\limits\limits_0^2 4xx^3 \; dx\]
 one year ago

jocelynevsq Group TitleBest ResponseYou've already chosen the best response.0
oh okay i see how you got that. thanks!
 one year ago
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