Calculus, all content (2017 edition)
- The fundamental theorem of calculus and accumulation functions
- Finding derivative with fundamental theorem of calculus
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Finding derivative with fundamental theorem of calculus
- Proof of fundamental theorem of calculus
How do you apply the fundamental theorem of calculus when both integral bounds are a function of x. Created by Sal Khan.
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- can someone explain why he added the times 2x at the back of F'(x)? I can't seem to understand that. I did however understand why he had to in the previous video. but in this case. i don't see the need.(14 votes)
- Because our upper bound was x², we have to use the chain rule to complete our conversion of the original derivative to match the upper bound. The derivative of x² is 2x, and the chain rule says we need to multiply that factor by the rest of the derivative.(26 votes)
- What would happen if you have intervals of numbers rather than x and x^2? For example if you have over the interval of 0 to 1 of the function cos(t)/t.(8 votes)
- Then, F(x) would be a constant since the input x is not used in the expression. If you take the derivative of a constant, the result is 0, meaning F'(x) would be 0.
I hope this helps!(42 votes)
- I don't really get how c is constant if its value depends on x. If by constant he means for a particular value of x, then wouldn't the bounds of the original integral also be constant and so wouldn't there be no point in splitting the integral into two parts in the first place?(13 votes)
- This is one of those confusing contraptions where the underlying function -- the one portrayed in the graph -- is a function of t, not a function of x. The constant c is a t-value, not an x-value. It doesn't depend on x. If you aren't getting how F(x) can be a function of x when we're doing an integral of a function of t, it may help to go back over the earlier videos in this section on the fundamental theorem.(12 votes)
- What about when the lower and upper limits of the integral both contain a variable for instance the integral from 3x to x^2 of 1/(2+e^t) ? How would you solve that problem?(2 votes)
- You simply do the integral in the normal way, and then substitute in the limits which are functions of x. You end up with an expression which is a function of x. This is quite reasonable, if you think about it -- a definite integral gives you the area below the curve between the two specified limits. If the limits depend on x, then the area is not going to be constant, but will also depend on x.
In your example we have
integral_(3x)^(x^2) 1/(2+e^t) dt
= [-ln(2+e^t) / 2 + t/2]_(3x)^(x^2)
= -ln(2+e^(x^2)) / 2 + (x^2)/2 + ln(2+e^(3x)) / 2 - 3x/2.(5 votes)
- At4:39Sal cancelled one x out of x^2 in the denominator with the only x in the numerator and wrote simply x but why can't x be negative?So i think it's reasonable to write ABSOLUTE VALUE OF X instead of simply x.Please help!(1 vote)
- Remember x is a variable, a placeholder for any value.
Now . . . .
Simplify x²/x. The answer is x. In no way have we limited x to be non negative.
In fact, by placing absolute values around the x, as you suggest, you are actually saying, "I don't care if the value x takes on is negative, make the result positive."
Example: let x = -5, then x²/x = (-5)²/-5 = (-5)(-5)/-5 = -5.
If we did it your way we would be saying that (-5)²/-5 = 5, and that is not correct.(6 votes)
- Instead of using c, would you be able to take the integral from 0 to x^2 - x ?(4 votes)
- I just solved it using 0 instead of c and it gave me the same answer, so I think it's acceptable.(1 vote)
- At4:21, why does Sal take the derivative of x^2? Didn't we already apply the Fundamental Theorem of Calculus, or does the fundamental theorem also state that we must take the derivative of the upper-bound?(2 votes)
- This video should help: https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-4/v/derivative-with-ftc-and-chain-rule
It has to do with the chain rule!(4 votes)
- Instead of thinking of a constant c between x and x^2, couldn't we also choose a constant, say k, whose value is lesser than x?
The solution would then be:
(derivative of integral from k to x^2)-(derivative of integral from k to x).
The results are the same, but then we don't need to switch the bounds.(2 votes)
- That depends. If the function is defined outside the interval given, (in this case x to x^2), which it should be, then yes, and you are correct in that the results would be the same. However, because the interval is given as x to x^2, it is considered "more correct" to use a value in between. If you have to show your work, I would do it this way, but otherwise, use whichever method works best for you.(3 votes)
So let's see if we can take the derivative of this expression right over here, if we can find capital F prime of x. And once again, it looks like you might be able to use the fundamental theorem of calculus. A big giveaway is that you're taking the derivative of a definite integral that gives you a function of x. But here I have x on both the upper and the lower boundary, and the fundamental theorem of calculus, is at least from what we've seen, is when we have x's only on the upper boundary. And then, of course, it's an x squared, but we've seen examples of that already when we used the chain rule to do it. But how can we break this up and put this in a form that's a little bit closer to what we're familiar with when we apply the fundamental theorem of calculus? And to realize that, we really just have to attempt to graph what this is representing. So let's say that this is our lowercase f of x, or I should say f of t. So let's call this lowercase f of t. And let's graph it over the interval between x and x squared. So let's say this is my y-axis. This is my t-axis. And let's say that this right over here is y is equal to f of t. I'm drawing it generally. I don't know what this exactly looks like. And we're going to talk about the interval between x and x squared. So if we're going to talk about the interval between x, which is right over here, it's the lower bound, so x and x squared. It's the lower bound, at least for this definite integral. We don't know for sure. It depends on what x you choose on which one is actually smaller. But let's just say that for the sake of visualizing, we'll draw x right over here, and we will draw x squared right over here. So this whole expression, this entire definite integral, is essentially asking for, is essentially representing this entire area, the entire area under the curve. But what we could do is introduce a constant that's someplace in between x and x squared. Let's say that constant is c, and break this area into two different areas with c as the divider. So that same exact whole area we can now write it as two separate integrals. So one integral that represents this area right over here, and then another integral that represents this area right over there, and where we just say c is some constant between x and x squared. Well, how can we denote this area in purple? Well, that's going to be-- So this thing is going to be equal to the sum of these two areas. The purple area we can show is the definite integral from x to c of our function of t, cosine t over t dt. And then to that we're going to add the green area. And then we'll get the original area. So for the green area, our lower bound of integration is now our constant c, and our upper bound of integration is x squared, and it's going to be of cosine t over t dt. And this is a form where, if we know how to apply the chain rule, we can apply the fundamental theorem of calculus. And this is almost in a form. We're used to seeing it where the x is the upper bound. And, well, we already know what happens. We can swap these two bounds, but it'll just be the negative of that integral. So this is going to be equal to-- let me rewrite it-- the negative of the definite integral from c to x of cosine t over t dt. And then we have plus the definite integral that goes from c to x squared of cosine t over t dt. So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus. So if we want to find F prime of x, well, applying the derivative operator over here, we're going to have a negative out front. It's going to be equal to negative cosine x over x. Once again, just the fundamental theorem of calculus. And then plus-- we're first going to take the derivative of this thing with respect to x squared, and that's going to give you cosine of x squared over x squared. Wherever you saw t, you replace it with an x squared. And then you're going to multiply that times the derivative of x squared with respect to x. So that's just going to be-- derivative of x squared with respect to x is just 2x. And we're done. We just need to simplify this thing. So all of this is going to be equal to negative cosine x over x plus-- well, this is going to cancel out with just one of those-- plus 2 cosine x squared over x. And I guess we could simplify it even more as being equal to-- and we can swap these-- everything over x 2 cosine of x squared minus cosine of x. And we are done.