AP®︎/College Calculus AB
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Taking derivatives and integrals of strangely defined functions. Created by Sal Khan.
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- At3:33we try to derive the integral evaluated from 0 to x. Would we not want to subtract the lower limit of f(0) from f(x)? Why is g'(x)=2+f(x) and not g'(x)=2+(f(x)-f(0))? This changes the answer since f(0) has a value.(7 votes)
- When you integrate, you get something like F(x) - F(0). F(x) is a function of x, while F(0) is an evaluation of F(x) at x = 0, which is some constant number. When you derive F(x) - F(0), d[F(x)]/dx = f(x), while d[F(0)]/dx = 0, since the derivative of a constant is zero. This equates to f(x) + 0 = f(x). Therefore, g'(x) = 2 + [f(x) - 0] = 2 + f(x).(13 votes)
- Why does the derivative of f(t) is f(x)?(2 votes)
- because of the first FTC. When you have an integral from a to x of some function with a different variable, in this case f(t) you simply replace it with x to make f(x) when you take the antiderivative.(3 votes)
- Why did Sal put -3 in f(x) when g'(-3) and f(x) are two different functions?(2 votes)
- The derivative of an integral from any number c to x (with respect to x) is just going to be that function at x. Basically, the derivative and integral simply cancel each other out, and since c is simply a constant it will disappear leaving just f(x), or in this case f(-3) since x is -3.(2 votes)
- Was anyone else tempted to find the equation of the circle (y^2 + x^2 = 9) and integrate it?
I always seem to miss the simple solution.(1 vote)
The continuous function f is defined on the interval negative 4 is less than or equal to x is less than or equal to 3. The graph of f consists of two quarter circles and one line segment. So this right here is one quarter circle, then we have another quarter circle, and then it has this line segment over here, as shown in the figure above. I put the figure on the side, so that we have some screen real estate down here. Let g of x be equal to 2x plus the definite integral from 0 to x of f of t dt. All right, that's a little bit interesting. Now, let's do Part a. Part a. Find g of negative 3. So we want to find g of negative 3. So first of all, it tells us what g of x is. So g of negative 3, every time we see an x, we just put a negative 3 there. So it's going to be 2 times negative 3 plus the definite integral from 0 to negative 3, f of t dt. So this first part is pretty straightforward. 2 times negative 3 is negative 6. And then this part right over here is the definite integral from 0 to negative 3 of f of t dt. So this is essentially the area under f of t-- or under f of x, if you want to view it that way-- from 0 to 3. And this is pretty straightforward to figure out but we have to be careful here. Because this area right over here, this would be the integral-- let me do this in a color that you can see. That area right over there would be the integral from negative 3 to 0 f of t, or we could even say f of t dt or f of x dx, either way. That would be this area right over here. They've swapped it. They have the larger number as the lower bound. They have 0 as the lower bound. So what we could do, we could rewrite this. This is the same thing, this is equal to negative 6 minus-- if you swap the bounds of integration, you swap the sign on the integral. So we could say minus from negative 3 to 0 of f of t dt. And now this thing right over here-- this expression right over here is the area under this quarter circle. And the area under this quarter circle, we can just use a little bit of geometry to figure that out. We know its radius. The radius here is 3. Radius is equal to 3. The entire circle is pi r squared. So the entire circle-- the area if this was an entire circle right over here-- that area would be pi r squared. So pi times 3 squared. So 9 pi. So it would be 9 pi, but this is only 1/4 of the entire circle. So we want to divide it by 4. So this right over here we'll evaluate to 9 pi over 4. So Part a, we get it's equal to negative 6, that's that right over there, minus 9 pi over 4. And that's at least the first part of Part a. Then they say find g prime of x. So let's find g prime of x. We did the first part of Part a right over there. Then we need to find g prime of x. Well, it's just going to be the derivative of g of x. The derivative of 2x is just going to be 2. And then the derivative-- and this is the fundamental theorem of calculus right over here-- the derivative of the definite integral from 0 to x of f of t dt, that's just going to be f of x. So that is g prime of x. We did this second part right over there. That is g prime of x. And then we need to evaluate it at negative 3. So g prime of negative 3 is going to be equal to 2 plus f of negative 3, which is equal to 2 plus-- and let's look at our-- let's see what f of negative 3 is. This is our function definition. When x is equal to negative 3, our function is at 0. So f of negative 3 is just 0. So it's just two plus 0, which is equal to 2. So g of negative 3 is equal to 2. And we're done with Part a. And the trickiest part of this is just realizing the bounds of integration. That you might be tempted to say, oh, the integral between 0 and negative 3, that's just this area over here because we're going between 0 and negative 3. But you've got to realize that we actually have to swap them and make them negative if you really want to consider this area right over here. That this integral is going to be the negative of this area. And that's what we did when we went through this step right over there.