AP®︎/College Calculus BC
- Particular solutions to differential equations: rational function
- Particular solutions to differential equations: exponential function
- Particular solutions to differential equations
- Worked example: finding a specific solution to a separable equation
- Worked example: separable equation with an implicit solution
- Particular solutions to separable differential equations
Sal finds f(0) given that f'(x)=5eˣ and f(7)=40+5e⁷.
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- At1:02, how is it allowed to 'take a constant out of the integral sign'?(8 votes)
- Since integral is in a way an infinite sum, 5 is constant with each term (it doesn't change) so you can just factor it out.(7 votes)
- What if you want to integrate an equation that has the variable in its exponent, and is also being multiplied with a constant?
Example:Integration of (0.67*e^0.044x)(3 votes)
- Well, you can take 0.67 out of the integral, then you can use u-substitution for e^0.044x, setting u =
0.044x. Then du = 0.044dx, or dx = du/0.044 = (1/0.044)du. Then the integral becomes 0.67∫(e^u)*(1/0.044)du. You can take 1/0.044 out of the integral since it is a constant. The integral of e^u is e^u. But you need to unsubstitute the u, so the answer is (0.67/0.044)*e^0.044x, or 15.227e^0.044x. Try taking the derivative of this to double check! Hope this made sense.(2 votes)
- I know the d(variable) in the antiderivative is used to let you know with respect to what you are antiderivating, but does it serve any other purpose? Or is there any other reason for it to be there?(2 votes)
- Good question. There is another reason, actually, and it stems from the early idea of an integral.
Remember when we treated the area under a curve as a summation of infinite, thin rectangles? Now, the area of a rectangle is length * width. The length of the rectangle was essentially the distance from y = 0 to the function. So, the length was the value of the function f(x). Now, the width was a small change in the x direction. So, we take it as Δx. We have the length and width. So, area of one rectangle = f(x)Δx.
Now, if the number of rectangles tends to infinity, Δx tends to 0. Whenever something tends to 0, Δ becomes d. So, area of a very, very small rectangle = f(x)dx. Now, as there are infinitely many, we integrate and hence, area under the curve = int(f(x)dx)
So, while dx gives you the variable of integration, it is also the width of a rectangle with infinitesimally small area.(2 votes)
- This is a bit of an off topic question, but I am learning integral calculus to comprehensively understand the Basel problem. So what should I target in order to comprehend the famous problem quicker? Thanks(1 vote)
- hi, one of the questions in the next test asks "dy/dx = 3y and y(0)=3. Find y(ln2)".
I'm quite confused about how dy/dx can be expressed in terms of y, rather than x. I'm not sure how to interpret it. Thanks.(1 vote)
- In cases where dy/dx has a y in it, the y itself would be a function of x. So, think of this as dy/dx = 3y(x). That should make more sense.
You can actually solve for y in this differential equation, differentiate that expression, substitute y in the DE given, and you'll see that they're equal.(1 vote)
- why not increase the power of e by 1....
in the case of 5e^x anti derivative ?(1 vote)
- Because e^x is an exponential function, and the power rule is not enough to evaluate its derivative. The power rule is applicable only to some variable in the base of a power and a constant in the exponent of that power. But with e^x the variable is in the exponent, and e (the base) is just constant number!
So we use the chain rule to differentiate e^x. If you're still not sure why, watch the lessons about the power rule and derivative of e^x(1 vote)
- hi guys after 7 years(0 votes)
- As Sal does not extract a value from the function f(7)= 40 + 5e^7, I can safely assume e is a variable and not the number e. We have two variables, x and e, and I am assuming a domain of all real numbers. At2:58, Sal says f(0)=5e^0 + 40 = 5(1) +40=45. Sal assumes e to be positive but he no where stated in defining his problem "for all e>0." Likewise, if e is a function of x and x's domain is all real numbers, the range is not inherently limited to positive values for e. My argument is that the answer is undefined as e^0 could be -1 or +1.(0 votes)
- In general, when you see e, it represents the base of the natural logarithm. Normally most people will never freely use e as a variable (nor should you), since it is normally assumed that e = 2.71828...
It's just one of those annoying math things that can seem ambiguous, but you are right, you should announce what you're using variables and constant symbols for. But doing so every time is also tedious and I doubt Sal wants to do it every time. Plus I believe he says later something about it being "just a value" so it should be insinuated from that, that he is using e as a constant, not a variable.(3 votes)
- [Voiceover] We're told that F of seven is equal to 40 plus five, E to the seventh power, and F prime of X is equal to five, E to the X. What is F of zero? So to evaluate F of zero, let's take the anti-derivative of F prime of X, and then we're going to have a constant of integration there, so we can use the information that they gave us up here that F of seven is equal to this. This might look like an expression. Well, it is an expression, but it's really just a number. There's no variables in this, and so we can use that to solve for our constant of integration, and then we will have fully known what F of X is, and we can use that to evaluate F of zero, so let's just do it. So if F prime of X is equal to five, E to the X, then F of X is going to be equal to the anti-derivative of F prime of X, or the anti-derivative of five, E to the X, DX, and this is the thing that I always find amazing about exponentials, and actually, let me just take a step. I'll take that five out of the integral so it becomes a little bit more obvious. And so the anti-derivative of E to the X, well, that's just E to the X because the derivative of E to the X is E to the X, which I find amazing every time I have to manipulate or take the derivative or anti-derivative of E to the X. So this is gonna be five, E to the X, plus C, and you can verify. Take the derivative of five, E to the X, plus C. The derivative of five, E to the X, well, that's five, E to the X, so that works out. Well, and the derivative of C is zero, so you wouldn't see it over here. So now, let's use this information to figure out what C is so that we know exactly what F of X is, and then we can evaluate F of zero. So we know that F of seven, so when X is equal to seven, this expression is going to evaluate to this thing, 40 plus five, E to the seven. So, five times E to the seventh power plus C is equal to 40, plus five, E to the seventh power. And notice, all I did is say, okay, F of seven. Well, if this is F of X-- Let me write this down. So, if this is F of seven, if this is F of X, I just replaced the X with a seven to find F of seven, and we know that F of seven is also going to be equal to that. They gave us that information, but when you just look at this, it's pretty easy to figure out what C is going to be. You can subtract five, E to the seven from both sides, and you see that C is equal to 40. And so we can rewrite F of X. We can say that F of X is equal to five, E to the X, plus C, which is 40. And so now, from that, we can evaluate F of zero. F of zero is going to be five times E to the zero power, plus 40. E to the zero is one, so it's gonna be five times one, which is just five, plus 40, which is equal to 45, and we're done.