# 5.4, 6.1 Area of Regions in the Plane

– Depending on the context that we’re working with, the area underneath the curve in a graph can represent the amount of money gained or lost or the amount of time lost or the efficiency of a product And so we’re often very interested in finding the geometric equivalent of the area under a curve in a plane So that’s going to be our question today is how do we find area under a curve? And the answer to that is that the definite integral is a representation of area Here’s what I mean by that If we’ve got some curve and we’re interested in the area between a and b that’s underneath this curve, that turns out to be the integral from a to b of our function, saying that function is f of x dx That is the way we find area is we take that definite integral between a and b So for example, if I want the area under x squared minus 10 between x equals 4 and x equals 6, what that means I’m going to find is the integral from 4 to 6 of that x squared minus 10 dx So we just need to evaluate that, and we know how to do that The antiderivative of x squared is x cubed times a 1/3 minus the antiderivative of 10 is 10x And we’re going to integrate that from 4 to 6 And so that just means we’re going to plug 6 in for the x So we have 6 cubed minus 4 cubed– since it’s a polynomial, we can just do the subtraction together– minus 10 times the 6 minus the 4 And if we plug that all in the calculator, we’ll end up with 92/3 for the total area between 4 and 6 underneath the x squared minus 10 So really what we’re doing is we’re just giving some context to what we learned how to do in the previous video Let’s say, for example, we want the area under f of x equals e to the negative x plus 2 that falls between x equals 1 and x equals 5 Well, area under a curve is an integral, and we’re integrating from 1 to 5 because that’s where I want the area to be of my function e to the negative x plus 2 dx So let’s evaluate this The antiderivative of e to the stuff is e to the stuff, and then we just have to divide by the derivative of negative x, which is negative 1 And we’re integrating this from 1 to 5 which means we plug those values in So we get negative e to the negative 5 plus 2, minus a negative, which makes it a plus, e to the negative 1 plus 2 Simplifying that becomes negative e to the negative 3 plus e to the first power And that gives us the area that’s under our curve between 1 and 5 Let’s try one more example Let’s find the area under the curve f of x equals e to the x times e to the x plus 2 cubed between x equals 0 and x equals 2 Well again, we know we have to integrate from 0 to 2 of our function e to the x times e to the x plus 2 cubed dx This one we can’t find the integral directly,

but we can use substitution if we make that e to the x plus 2 equal to the u That way it becomes u cubed And we can do that because its derivative is multiplied in there of e to the x dx So now we have the integral from– make sure we plug these limits into our u equation So we’ve got e squared plus 2 It’s a funny looking limit, but it’s a limit If we plug 0 in, we get e to the 0, which is 1, plus 2 is 3 So we’re integrating from 3 to the e squared plus 2 The e to the x dx all becomes du The rest becomes u cubed And we know we can increase the exponent by 1 and divide by the new exponent– 1/4 u to the fourth, integrated from 3 to the e squared plus 2 limit Plugging those limits in, we’ll end up with 1/4 times e squared plus 2 raised to the fourth power minus 1/4 times 3 to the fourth power And simplifying that, we end up with 1/4 e squared plus 2 to the fourth power minus 3 to the fourth is 81/4 So we end up with our solution– the area under the curve– which sounds pretty straight forward But there is one little problem that comes up that we need to address A problem and a solution And it’s just something we need to be careful of when we’re talking about area under a curve Let’s take a look at the area under f of x equals x squared minus 4 between x equals 2 and x equals negative 2 Well, we know to do this we just have to integrate from our limits The lower limit, the smallest number’s negative 2 up to 2 of our function, which is x squared minus 4 dx And we can take that antiderivative quick enough at this point That’s x cubed times 1/3 minus 4x And we know we want to integrate from negative 2 to 2 So we plug those limits in 1/3 times 2 cubed minus 4 times 2 And then we subtract, which will change the signs, plugging the negative 2 in 1/3 times negative 2 cubed Changing the sign, now we have plus 4 times negative 2 Plug all that in, and we’ll end up with an area of negative 32/3 How is area negative? Area can’t be negative Something’s wrong In fact, the problem can get even worse Let’s take a look at a problem like this where I want to find the area under f of x equals x cubed between x equals negative 2 and x equals 2 Well, using our old methods, we have the integral from negative 2 to 2– negative 2’s the small, 2’s the big– of our function x cubed dx which is 1/4 x to the fourth integrated from negative 2 to 2 So we plug the top value in– 2 to the fourth times 1/4 minus– plugging the bottom value in– 1/4 times negative 2 to the fourth And if we do that math, we end up with 0 How is there zero area? There is no area between negative 2 and 2?

That doesn’t make sense Let’s see if we can get a visual idea of the reason behind what’s happening with these two graphs, and maybe that reason will help us find a solution Our first function was f of x equals x squared minus 4 And we know x squared minus 4 is a parabola that’s shifted down 4, and actually has x-intercepts of negative 2 and positive 2 So when we found the area between negative 2 and positive 2, that entire area covered all those negative y values It’s below the y-axis, making it a negative value And similarly, when we did f of x equals x cubed– we should be familiar with the graph of x cubed as one that comes up, levels off, and takes off, and it’s perfectly symmetrical So when we go from negative 2 to positive 2, we end up with a positive area and a negative area connected to the x-axis And that negative area is exactly the same size as the positive area, which gives us a combined total of 0 But that doesn’t mean those areas are 0 and negative We need a better way to attack these And so our solution is going to be to subtract the negative parts In other words, when we did the problem– the first example, where we found the area under x squared minus 4 between x equals negative 2 and 2, looking above, we see that graph is entirely negative And so what we’ll do is we’ll add an extra negative, subtracting that integral from the negative 2 to 2 of our function x squared minus 4 dx That negative there is going to change the entire thing to a positive Basically, it gives us the other part of the graph shifting the other direction, which now becomes positive And that negative sign is going to make all the difference in the world because now when we take the antiderivative, we get the opposite of x cubed times 1/3 minus 4x integrated from negative 2 to 2 And now when we plug in the lower limit, we get the negative 1/3 times 2 cubed minus 4x– not 4x 4 times 2– plugging in the upper limit– minus the negative, which is now positive 1/3 times negative 2 cubed minus 4 times negative 2 And now when I put that value in the calculator, we’ll end up with a positive 32/3, and that’s a much more reasonable value being positive for the area The second example, though, we need to be a little bit more clever with because what we’re looking for here is the area– we’ve got that negative area– looking at the graph above– to the left, and the positive area to the right So we’re going to treat that like two separate integrals Notice, it switches at 0, 0 So we’re going to do the left side from negative 2 to 0, and we’re going to subtract that integral So again, we’re finding the area under x cubed between x equals negative 2 and x equals 2, and because the first part is negative, we’ll put a negative sign again So a negative negative makes a positive But we’re only going to integrate from negative 2 up to where the negative part stops, which is 0, of our function x cubed dx

The next part is already positive, so we’ll have a positive integral from that changing point up to our limit of 2 of our function x cubed dx And that will give us our total area, all positive, because we took the negative part and flipped it, taking the opposite sign So let’s find this solution We’ve got negative 1/4 x to the fourth, integrated from negative 2 to 0, plus a 1/4 x to the fourth integrated from 0 to 2 Plugging in those values, we have negative 1/4 times 0 to the fourth Minus a negative makes it a positive 1/4 times negative 2 to the fourth Plus, plugging the limits into the right half, plugging first 2 in– 1/4 times 2 to the fourth minus the lower limit– 1/4 times 0 to the fourth And now, if we work this out, the 0’s go away pretty quick That’s nice 2 the fourth is 16, divided by 4 is 4 So this becomes 4 plus 4, for a total area of 8 And now, instead of 0, we have a more reasonable area of 8 What we’ve been doing so far has been looking to find the area underneath a curve, like we did in our introduction where we said, hey, all the area goes down to the x-axis But sometimes we don’t want to go all the way down to the x-axis We only want to go down to another function Sometimes we want to find the area between two curves The idea here for the area between two curves is we have one curve up above and one curve down below, and we want the area between a and b that’s between those two values Well, if we were to integrate f of x from a to b, that would give us all the area underneath f all the way down to the x-axis But we don’t want all of it because we don’t want the stuff that’s below the second graph We’ll call that g of x We don’t want any of that lower area Well, we know we can calculate that lower area by integrating g of x dx over the area from a to b, but that’s the area we don’t want And so what we do for the part we don’t want is we’re going to subtract off that area When we subtract that area, we end up getting only the area between the two curves So again, the way we find the area between two curves is we integrate from a to b and subtract the bottom function from the top function Let’s see if we can take a look at an example where we do that, and we’re going to use our calculators a bit to help us identify which one exactly is the top function that needs to come first So we’re going to start by finding the area between f of x equals x plus 5, g of x equals x squared minus 2x plus 2, and x equals 0 and x equals 3 Well, first we need to know which graph is on top Going to my calculator, if I hit y equals, we can type in both these equations x plus 5 is the first equation, and x squared minus 2x plus 2 is the second equation Now, when I hit Graph, I’m going to watch the graph be drawn because the top equation is going to be drawn first There’s the top equation And there’s the bottom equation Notice, we’re only interested in going from 0 to 3

So between 0 and 3, what we notice is the line is on top and the parabola is on the bottom between 0 and 3 So we’re going to set up our integral just like that We’re going to integrate from 0 to 3 the top function, which we said was the line x plus 5 Then we will subtract the bottom function, which we said was the x squared minus 2x plus 2 dx This integral will then give us the area between the two curves Let’s simplify first by distributing that negative through the parentheses, giving us x plus 5 minus x squared plus 2x minus 2 dx, and combining like terms to get negative x squared plus 3x plus 3 dx Evaluating is pretty straightforward We’ve got x cubed times negative 1/3 plus x squared times 3/2 plus 3x integrated from 0 to 3 Plugging those values in, then, we’ve got negative 1/3 times 3 cubed plus 3/2 times 3 squared plus 3 times 3 And then we subtract or change the signs as we plug the lower limit in The negative becomes positive 1/3 times 0 cubed minus 3/2 times 0 squared minus 3 times 0 And I’d probably do that all on my calculator to end up with 27/2 is the area that’s between my two curves Let’s try one more example before we wrap up Let’s find the area between 2x minus 3 and negative x squared plus 4x We’ll go into the calculator Hitting y equals, we’ll clear out the two functions The first function is 2x minus 3 The second function– negative x squared plus 4x And when I hit Graph, there’s the line There’s the parabola And we’re looking for this area in between right here Well, one thing we need to know is where they intersect We’re going to have to solve this equation for where they intersect, and then the parabola, the second function’s on top So first, to figure out where they intersect– those are the limits– they intersect when the two equations cross each other– when 2x minus 3 is equal to the negative x squared plus 4x Well, if we add x squared to both sides and subtract 4x, keeping the minus 3, we can factor that to x minus 3x plus 1 So they intersect at 3 and negative 1 So that tells me that we’re going to integrate from negative 1 to 3, from low to high of the top function we said was the squared, so negative x squared plus 4x minus the bottom function And the bottom function was the 2x minus 3 dx And now we have an integral we can solve to find the area between the two curves Simplifying by distributing the negative through We have negative x squared plus 4x minus 2x plus 3 dx Combining like terms, we have negative x squared plus 2x plus 3 dx Taking the antiderivative, we end up

with negative x cubed times 1/3 plus x squared divided by 2 clears the 2 plus 3x integrated from negative 1 to 3 So we plug in our values Plugging the top one in, negative 1/3 times 3 cubed plus 3 squared plus 3 times 3 Subtracting, changing the signs, we now have a plus 1/3 times negative 1 cubed minus negative 1 squared and minus 3 times negative 1 And if we plug that all into the calculator, we’ll get the area between the two curves which comes out to be 32/3 So area really is just a major application of that definite integral, finding the amount of stuff between the values Be careful if you’ve got a negative area We’ll have to figure out how to put the negative sign in front to make it actually positive area so we don’t end up with 0 or negatives Also, when we want the area between two curves, we just need to subtract the two curves– one from the other So take a look at the homework assignment to practice some of these We will see you in class to discuss applications and practice these further