Two equal areas

Added: 2026-05-18

[00:00:05]Here is a question.

[00:00:11]The diagram below here. The diagram below. The diagram below.

[00:00:21]shows this a graph of the function f of x equals x.

[00:00:30]No.

[00:00:31]f of x equals e to the negative x power.

[00:00:39]This is a function f of x equals e to the negative x.

[00:00:49]This is the graph.

[00:00:53]When x is zero, e to the power zero is one. So, this is one.

[00:01:01]This is 0.5.

[00:01:05]This is 1.5.

[00:01:09]The function looks like this. It's decreasing.

[00:01:15]When x is larger, the function the graph is decreasing.

[00:01:25]Another function is g of x.

[00:01:31]It is log x + 1.

[00:01:37]When x is zero, this is just log one. Log one is zero. So, when x is zero, the function g is zero.

[00:01:51]This is the graph.

[00:01:54]The graph of g when X is zero the function G is zero.

[00:02:02]And this log function is increasing like this.

[00:02:11]So, one is decreasing.

[00:02:15]Another one is increasing.

[00:02:18]They intersect at the point P.

[00:02:26]Now, what does the question want?

[00:02:32]The graphs intersect at P.

[00:02:39]And the regions enclosed by the graph from X equals zero to X equals K are shaded. So, the region from X equals zero and X equals K are shaded. The region here shaded. The region here is shaded.

[00:03:15]The area of the shaded region left of P.

[00:03:24]This is P.

[00:03:26]The shaded region to the left side of P, which is this.

[00:03:32]This shaded region The region left of P is equal to the to the area of the shaded region to the right-hand side of P.

[00:04:01]This shaded region to the left of P, the area is equal to this area to the right-hand side of P.

[00:04:16]So, the question wants this area is equal to this area.

[00:04:24]What is the value of K?

[00:04:27]>> [clears throat] >> Find the value of K.

[00:04:37]So, this question is illustrated by the graph.

[00:04:45]I have a decreasing function.

[00:04:49]I have an increasing function.

[00:04:53]They intersect at P.

[00:04:57]And at what value of K I have this area equals this area. Find the K.

[00:05:10]So, what is the way of doing this?

[00:05:15]So, we will think.

[00:05:18]I have the given function F.

[00:05:24]I also have the given function G.

[00:05:28]They intersect. Therefore, I can let this function equals this log function and solve for X.

[00:05:40]And this value of X will give me P.

[00:05:47]Will give me coordinates of P.

[00:05:50]Once I have coordinates of P, I can find the area using integration.

[00:06:00]This function minus this function integrate from zero to the X coordinate I just found, right?

[00:06:12]The X coordinate of intersection. So, I do that, I find the numerical value of the area.

[00:06:22]Then I write down another integration.

[00:06:26]This function the G function minus F function integrate with respect to X from the X value of P which we already found from the X value of P to the unknown X value K.

[00:06:51]And then this area is unknown because I don't know K.

[00:06:58]But this area I already found.

[00:07:01]So, I let this area equals the area I already found and I have an equation and I can solve for K.

[00:07:14]So, that is the idea.

[00:07:17]However, this is quite long because first you have to find the intersecting point P.

[00:07:28]Let's do it another way.

[00:07:43]>> [snorts] >> So, question we want to do is area to the left of P equals area to the right of P.

[00:07:56]So, left of P is the area is F minus G because F is above G.

[00:08:05]So, F minus G integrate from X equals 0 to X equals P.

[00:08:13]This area is equal to the area to the right-hand side of P.

[00:08:18]On the right-hand side, the function G is above the function F.

[00:08:25]So, I have G minus F integrate from P to K.

[00:08:31]So, these two areas are equal.

[00:08:37]I move this right-hand side expression to the left-hand side.

[00:08:44]Then, I have a negative here.

[00:08:52]Move this to the other side, I have a negative. So, that's that's this next step.

[00:09:00]Now, I can do a trick.

[00:09:04]I move the negative here to the inside of integration.

[00:09:12]So, the negative becomes plus becomes positive and I have negative G positive F. So, which means I move the negative sign after the integral sign.

[00:09:29]Then, I have plus F minus G. F minus G.

[00:09:37]What is the advantage of doing this step?

[00:09:42]Now you see the integration is f minus g.

[00:09:48]f minus g and the the limit of integration is from zero to p plus from p to k.

[00:09:59]>> [clears throat] >> Do you remember that these two integrals [clears throat] can be combined because from zero to p plus from p to k is the same as from zero to k.

[00:10:22]because the integral is the same f minus g f minus g.

[00:10:31]So I have f minus g from zero to k and this is equal to zero.

[00:10:38]Let me put in the function. f of f is e to the negative x g is logarithm x plus one.

[00:10:56]I separate the integration into two parts. The first part is just this. The first part is this.

[00:11:07]The second part is this.

[00:11:12]The first part you can integrate.

[00:11:18]You get a negative sign here because you have a negative here. So the integration of this is negative e to the power negative x.

[00:11:30]The limits The limit of integration is from zero to K.

[00:11:35]And this one you need to use integration by parts.

[00:11:42]This is U.

[00:11:44]This is DV.

[00:11:46]UDV integration is UV minus VDU.

[00:11:54]So, this is equal to UV. U times V. So, U times V. Minus VDU.

[00:12:06]V is this, X. V. DU.

[00:12:11]D in logarithm is this. This is DU. So, VDU. [snorts] I rewrite this part into 1 minus 1 over X plus 1.

[00:12:33]Then I can integrate.

[00:12:39]Uh this part is easy.

[00:12:45]This part is this. All right? This is easy.

[00:12:49]Uh I almost miss a negative sign. So, there is a negative X here. So, you put in the limits and I have this.

[00:13:02]Here, this one you put in the limits.

[00:13:06]I have this. Okay.

[00:13:11]This one you integrate each part. This is 1 DX and minus this DX. So, you integrate this, you have this result. And is upper limit and lower limit and so on. So, you put in the limit I have this.

[00:13:35]Now, I need to solve this equation for K.

[00:13:46]What K value makes this equal to zero?

[00:13:51]This is very difficult to find.

[00:13:54]Therefore, I need to use Desmos Desmos graphing calculator.

[00:14:04]So, use the Desmos graphing calculator to graph the left-hand side function. To graph that.

[00:14:15]And what value of K the function is zero? That means you find X intercept.

[00:14:23]So, I graph it.

[00:14:31]This is my graph.

[00:14:34]And the X intercept you have X equals zero and X equals this.

[00:14:41]But I don't want X equals zero.

[00:14:44]I want this value of X. So, the graphing calculator tells me K is this.

[00:14:52]This is how you solve this problem.

[00:15:00]Okay.

[00:15:01]Algebraically, I have difficulty to solve the value of K. So, the last step I need to use a graphing calculator.

[00:15:15]So, this is the end of this problem.

[00:15:18]It's a little longer. We You a graphing calculator.

[00:15:23]That's it.

[00:15:25]Thank you for watching.

[00:15:27]See you next time.

[00:15:30]Goodbye.