We can use a definite integral in terms of 𝘺 to find the area between a curve and the 𝘺-axis.
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- Why do you have to do the ln of the absolute value of y as the integral of a constant divided by y? In other words, why 15ln|y| and not 15lny?(6 votes)
- Because logarithmic functions cannot take negative inputs, so the absolute value sign ensures that the input is positive.(12 votes)
- In the video, Sal finds the inverse function to calculate the definite integral. What if the inverse function is too hard to be found? Is there an alternative way to calculate the integral?(8 votes)
- Bit late but if anyone else is wondering the same thing, you will always be able to find the inverse function as an implicit relation if not an explicit function of the form y = f(x). From there on, you have to find the area under the curve for that implicit relation, which is extremely difficult but here's something to look into if you're interested:
- why are there two ends in the title? Typo?(6 votes)
- Would finding the inverse function work for this? I know the inverse function for this is the same as its original function, and that's why I was able to get 30 by applying the fundamental theorem of calculus to the inverse, but I was just wondering if this applies to other functions (probably not but still curious).(3 votes)
- The way I did it initially was definite integral 15/e^3 to 15/e of (15/x - e)dx + 15/e^3(20-e) I got an answer that is very close to the actually result, I don't know if I did any calculation errors(1 vote)
- Can you just solve for the x coordinates by plugging in e and e^3 to the function? It seems like that is much easier than finding the inverse.(0 votes)
- how can I fi d the area bounded by curve y=4x-x and a line y=3(0 votes)
- [Instructor] So right over here, I have the graph of the function y is equal to 15 over x, or at least I see the part of it for positive values of x. And what I'm curious about in this video is I want to find the area not between this curve and the positive x-axis, I want to find the area between the curve and the y-axis, bounded not by two x-values, but bounded by two y-values, so with the bottom bound of the horizontal line y is equal to e and an upper bound with y is equal to e to the third power. So pause this video, and see if you can work through it. So one way to think about it, this is just like definite integrals we've done where we're looking between the curve and the x-axis, but now it looks like things are swapped around. We now care about the y-axis. So let's just rewrite our function here, and let's rewrite it in terms of x. So if y is equal to 15 over x, that means if we multiply both sides by x, xy is equal to 15. And if we divide both sides by y, we get x is equal to 15 over y. These right over here are all going to be equivalent. Now how does this right over help you? Well, think about the area. Think about estimating the area as a bunch of little rectangles here. So that's one rectangle, and then another rectangle right over there, and then another rectangle right over there. So what's the area of each of those rectangles? So the width here, that is going to be x, but we can express x as a function of y. So that's the width right over there, and we know that that's going to be 15 over y. And then what's the height gonna be? Well, that's going to be a very small change in y. The height is going to be dy. So the area of one of those little rectangles right over there, say the area of that one right over there, you could view as, let me do it over here, as 15 over y, dy. And then we want to sum all of these little rectangles from y is equal to e, all the way to y is equal to e to the third power. So that's what our definite integral does. We go from y is equal to e to y is equal to e to the third power. So all we did, we're used to seeing things like this, where this would be 15 over x, dx. All we're doing here is, this is 15 over y, dy. So let's evaluate this. So we take the antiderivative of 15 over y and then evaluate at these two points. So this is going to be equal to antiderivative of one over y is going to be the natural log of the absolute value of y. So it's 15 times the natural log of the absolute value of y, and then we're going to evaluate that at our endpoints. So we're going to evaluate it at e to the third and at e. So let's first evaluate at e to the third. So that's 15 times the natural log, the absolute time, the natural, (laughs) the natural log of the absolute value of e to the third power minus 15 times the natural log of the absolute value of e. So what does this simplify to? The natural log of e to the third power, what power do I have to raise e to, to get to e to the third? Well, that's just going to be three. And then the natural log of e, what power do I have to raise e to, to get e? Well, that's just one. So this is 15 times three minus 15. So that is all going to get us to 30, and we are done, 45 minus 15.