Fundamental Theorem of Calculus: Study with Video Lessons, Practice Problems & Examples

The fundamental theorem of calculus connects definite integrals and antiderivatives, allowing for efficient evaluation of integrals. Part one states that if a function is continuous on an interval [a, b], the derivative of its antiderivative equals the original function: $F^{\text{'}}=f\left(x\right)$. Part two enables evaluation of a definite integral by calculating the antiderivative at the bounds: $\int f_{\left(}dx=F\left(b\right)-F\left(a\right)$.

Fundamental Theorem of Calculus Part 1

In recent videos, we've spent some time talking about definite integrals. We've discussed what they represent and the various rules that apply. Now, what we're going to take a look at in this video is a very important use for definite integrals, which is known as the fundamental theorem of calculus part one. If that sounds super important, it is. But don't sweat it because in this video, I'm going to walk you through some examples that we will likely see in this course, and I'm also going to show you how surprisingly simple this equation and theorem is once you get the hang of it.

So without further ado, let's get right into things. I want you to recall something. We've talked about antiderivatives in the past, and recall that the function big F of x is the antiderivative of the function little f of x. The reason I'm asking you to recall this is that it's going to be important for understanding the fundamental theorem of calculus. For the fundamental theorem of calculus part one to work, a couple of things need to be true. We need to have a function that is continuous on some interval a to b. We need our function to be continuous so there are no breaks or jumps. And we also need this integral to exist. And if this integral exists, there's something we can say about its derivative because that's what the fundamental theorem of calculus part one is all about. This theorem shows us the connection between derivatives and definite integrals.

So, focusing on our equation, the main fundamental theorem says this: if we take the derivative of big F of x, which we recall in the past has been the antiderivative, then what we're essentially doing is taking the derivative of this definite integral. So we have this definite integral that goes from some number a, to x of this function here, and what we're going to say is that this derivative is going to come out to little f of x. If this looks confusing at first, I want you to focus on something: what we're literally saying is if we take the derivative of the antiderivative, we get just the function itself. And that should make sense because, basically, all this is saying is that the derivative and antiderivative will cancel each other out.

Notice we're taking the derivative of an integral. But we've seen with indefinite integrals that in essence, the process for this integral is just like the reverse, or the antiderivative. So what we can see here is that this derivative is basically canceling this integral and giving us just the function. One main thing that does change here is this variable. Notice we had a t inside, and now we have an x. The reason for that is because we're taking this function and bounding it from some random number a all the way up to x. So it changes everything to x, but that's okay because when we take this derivative, we just give the same function. We just switch up the t's with x's.

Now we'll start with example a. Notice for example a, we have this equation here, y is equal to this integral. And for both of these examples, we're asked to use the fundamental theorem to find dy/dx for the following, which is the derivative. So if I want to find the derivative of this integral, that's going to give me the solution to my problem. But what I can do is use the fundamental theorem. And that says that I can just take these t's and replace them with x's because this is just going to be the same function we have. So dy/dx is going to be x cubed minus three x squared plus four, and that right there would be the solution. Notice how quick it is to use the fundamental theorem of calculus. We didn't have to do this long and painful process here to find this derivative.

We simply just switched the variable t with x. And remember, that's all our fundamental theorem says. We just take our function t that's inside the integral and replace it with x, and that would give us the solution. But now let's try example b. And example b is a little more tricky because notice that we're going from some value of five all the way up to x squared. We no longer have an x on top of our integral. Now we have an x squared. So what exactly do we do when this happens? Well, whenever we have a situation where the upper bound of our integral is a function of x, what we want to do is use the fundamental theorem of calculus with the chain rule. The chain rule involves taking the derivative of the outside and multiplying it by the derivative of the inside to get the total derivative for our function.

And if we want to use the chain rule when dealing with the fundamental theorem, it looks like this: we take whatever this t is and replace it with this function of x on top, just like we had up here. The main difference, though, is now you have to take this entire thing and multiply it by the derivative of the top function. So what I need to do first to find dy/dx is to first take this t and replace it with x squared. That's taking this function and just putting it into what we have. So that would give us two over one plus x squared. Now what I need to do from here is multiply this by the derivative of the top function, and the derivative of x squared with respect to x is two x. Now from here, I can multiply these twos to give me four x on top and then one plus x squared on the bottom, and that is the solution to example b. It's pretty quick to solve these problems when we use the fundamental theorem of calculus. And the main thing you should remember from this video, if you're just looking at these equations, is to remember how to use this one.

Because notice here that I could use this exact same equation for the first example that we tried. Because if I went ahead and took this function of x and replaced it for t, that would give me what we had down here, and then I could multiply this by the derivative of the outside function or the top function here. But notice that would just be the derivative of x, which is one. So we would just get the exact same thing that we did if we used the initial theorem that we learned about. But the main thing that we need to remember is this scenario where if you take the derivative of an integral, you're just replacing the inside of that function with the top here, and you may need to apply the chain rule if there's a function of x on top that is not just x.

That is the idea for the fundamental theorem of calculus part one. I highly encourage you to check out the practice and examples after this video to ensure we have this concept down. See you in the next one.

Given the definite integral $F\left(x\right)={\int }_3^x\text{\hspace{0.17em}⁣}[t^8-\sin \left(t^4\right)]dt$ , find the derivative $F^{\prime }\left(x\right)$.

• A. $F^{\prime }\left(x\right)=3x^9-3x\sin \left(x^4\right)$
• B. $F^{\prime }\left(x\right)=x^9-x\sin \left(x^4\right)$
• C. $F^{\prime }\left(x\right)=3x^8-3\sin \left(x^4\right)$
• D. $F^{\prime }\left(x\right)=x^8-\sin \left(x^4\right)$

Answer: (D)

Given the definite integral $F\left(x\right)=\int_{12}^{20x}\!\left(h^4+\frac{63h}{\sqrt{h^5}}\right)\,dh$, find the derivative $F^{\prime}\left(x\right)$.

• A. $F^{\prime }\left(x\right)={20}^5{x}^4+\frac{25,200x}{\sqrt{{\left(20x\right)}^5}}$
• B. $F^{\prime }{\left(x\right)=({20}^{}x)}^4+\frac{1260x}{\sqrt{{\left(20x\right)}^5}}$
• C. $F^{\prime }\left(x\right)={\left(20x\right)}^5+\frac{25,200x^2}{\sqrt{{\left(20x\right)}^5}}$
• D. $F^{\prime }\left(x\right)={\left(20x\right)}^5+\frac{1260x^2}{\sqrt{{\left(20x\right)}^5}}$

Answer: (A)

Fundamental Theorem of Calculus Part 1 Example1

Let's go ahead and try this problem. Here, we're asked to find the derivative big f prime of x of the following definite integrals. Notice in these problems we have these pretty long definite integrals that we're dealing with, and normally having to compute these integrals and their derivatives respectively would be a huge pain. However, we learned that the Fundamental Theorem of Calculus Part One allows us to use a shortcut for these types of problems where we're taking the derivative of something being integrated. So let's see if we can apply that fundamental theorem to solve these problems in a more efficient fashion.

First, we're going to start with this one. We're trying to find the derivative big f prime of x. Notice that our upper bound is x. What we've said for the Fundamental Theorem of Calculus is if we're taking the derivative of something that is being integrated, the derivative and integral will, in essence, cancel each other out. So, all we're going to have to do is take this upper bound and replace it wherever we have the other variable, which in this case is t.

Replacing t with x, we're going to have x4−30x3+x23+30x−1000. And that right there would be the result and the solution to the problem. See? It's very quick to use the Fundamental Theorem, and as you can see, it allows us to solve these kind of complex problems in a quick and simple way. Now let's move on to example b.

We did example a. Now let's check out the second example here. Here we are asked to take the derivative of this integral. Now, this one's a little trickier because we have this function of x on top. It's not just a straight x by itself.

This means we're going to need to find a slightly different strategy to solve this problem, but we can still use the Fundamental Theorem of Calculus. When you take the derivative of an integral and the upper bound is a function of x, well, recall that there's a strategy we can use. We first take this upper bound of x and substitute it for every place we have the new variable, which in this case is h. So in this case, what we'll have as our derivative here is equal to x2−13x8x2−13x. Then, it's going to be minus, and then we'll have the sine of x2−13x. So this is what we get. However, this is not the final solution to the problem right here because we have this function of x on top. Whenever you have a function of x on top, you need to use the chain rule to solve the problem.

Using the chain rule here, we need to take this whole expression and multiply it by the derivative of the upper bound. The derivative of x2−13x is going to be the derivative of x², which is 2x minus the derivative of 13 with x, which is just 13. Whenever you do this, you need to take the derivative of whichever bound has the function of x. It's often the upper bound, but whatever is the function of x, you take the derivative of, multiply it on the outside and that's going to give you the result for big f prime of x. Now, I am going to simplify this a little bit more just so we don't have this 2x - 13 kind of dangling on the outside here.

So, I'm going to take this whole function that we have, and I'll go ahead and rewrite it down here. So this is our function. And what I'm going to do is see if I can rearrange things, so it's a little bit nicer looking. So we have big f prime of x and we'll go ahead and move this piece right here. It'll just take this 2x - 13 and bring it up to the top.

So we have 2x−13. And this is a little bit more aesthetically pleasing to look at. So, this will be our solution. This right here will be the solution to our problem. This would be the solution to example a.

And that is how you can solve these problems using the Fundamental Theorem of Calculus Part One. Hope you found this video helpful.

Fundamental Theorem of Calculus Part 2

Welcome back, everyone. So in the last video, we got introduced to the fundamental theorem of calculus part one, and this gave us a relationship between definite integrals and antiderivatives. Now what we're going to take a look at in this video is the fundamental theorem of calculus part two, and this actually gives us a straightforward way of dealing with these definite integrals when we see them. So, without further ado, let's take a look at what this theorem is, and hopefully, this is going to seem very straightforward after I walk you through it and show you some examples. So, let's get into this.

Now, when it comes to the fundamental theorem of calculus part two, this allows us to evaluate a definite integral using the antiderivative at the upper and lower bounds. So an example would be this situation down here. If we have this integral that goes from a to b, there are a couple of things that need to be true. First off, f(x) needs to be continuous on this interval from a to b, and then big F(x), the antiderivative of this function down here, is any antiderivative of this function, little f(x), on the same interval. If those are true, then we can say this.

Our integral from a to b of our function is going to be F b − F a where F is the antiderivative of f. So what we're, in essence, doing is finding the antiderivative of this function and then plugging in the high and low bounds, subtracting our results, and that will give us the solution to our definite integral. So, let's go ahead and solve an example down here.

Now, we'll first take a look at example a. Notice example a asks us to deal with this definite integral. And to evaluate it, well, let's use this fundamental theorem of calculus. What we're going to do is first find the antiderivative of 2x. Now, I can use the power rule here, where I take this exponent, add one to it, and then divide by the new exponent, and doing that is going to give me x². So this here is going to be the antiderivative, and we're going to bound this from our low bound of two to our high bound of five.

So this bar that we see right here, this is just another way of saying that we've already taken this antiderivative. But what we now need to do is we need to plug in the five and the two and evaluate. Now, you might notice that I did not add a plus c after this function, and I'll explain why I didn't do that in a moment here. So what we're going to do is just use this value that we got after taking the antiderivative, and we're going to take our five and plug it in. So that's going to be five squared, and we're going to take our two and plug it in. That's going to be two squared. And according to the fundamental theorem of calculus part two, we need to subtract these results. So we get 25 minus 4, which is going to come out to 21. So 21 is the solution to this problem.

Now, why didn't I add the plus c earlier? Well, the reason for that is notice how we ended up just subtracting these two results. If I plugged in a plus c into this function, the two plus c's would have just canceled each other out when we subtracted. So there is no point in putting the plus c when dealing with these definite integrals. So you don't need to worry about the +c's if you have bounds on your integral.

But now let's take a look at another more complicated example, example b. In this example, notice that we're asked to find this integral. Now, the way that I'm going to solve this problem is by first taking the antiderivative. And the way that I can do this is by going over each one of these terms and taking the antiderivative individually. We learned that that was one of the rules when we have a bunch of terms being subtracted or added. So I'm first going to use the power rule on this first term, and that's going to give me x³ over three. Then we're going to have minus, and then using the same power rule on this term we're going to have 2x². Then we're going to have plus, and I'm going to take that five and slap an x on there that gives me 5x. And then we need to bound this whole thing that we have right here from one to four.

Now, next, I can actually write this more simplified. So we have x³ over three, we're going to have minus 2x², and we're going to have plus 5x. And this whole thing is still bounded from one to four. Now, what I need to do next is I need to plug in my values and subtract. So I'm first going to plug in the values that I have for four into this function. So I plug in four, I'm going to get (64/3) minus 32 plus 20, which gives us 44/3 for the upper bound. And then for the lower bound of one, we have (1/3) - 2 + 5, which results in 11/3. Subtracting these, we end up with 33/3 or 11 as the final result. So 11 would be the solution to our problem.

So if we had this function that was sketched on a graph, the area underneath the curve of that function would be 11 on this interval from one to four. And for this function that we had over here, the area underneath this function from two to five would be 21. And that's exactly what the definite integral tool does in both of these examples. So I hope you found this video helpful, and we're going to have a lot of practice after this video so we can become very familiar with using this part two of the fundamental theorem of calculus. See you in the next one.

Evaluate the following integral:
${\int }_0^3\text{\hspace{0.17em}⁣}4dx$

• A. 12
• B. 4
• C. 3
• D. 0

Answer: (A)

Evaluate the following integral:
${\int }_1^4\text{\hspace{0.17em}⁣}(2x+1)dx$

• A. 16
• B. 24
• C. 12
• D. 18

Answer: (D)

Evaluate the following integral:
${\int }_{-1}^2\text{\hspace{0.17em}⁣}(x^2-3x+2)dx$

• A. -6
• B. 0
• C. $\frac{27}{6}$
• D. $\frac{25}{6}$

Answer: (C)

Evaluate the following integral:
${\int }_0^1\text{\hspace{0.17em}⁣}(2x^3-x^2+4x)dx$

• A. $\frac{13}{6}$
• B. $\frac{1}{6}$
• C. $\frac{17}{6}$
• D. $-\frac{11}{6}$

Answer: (A)

Evaluate the following integral:
${\int }_2^3\text{\hspace{0.17em}⁣}{x}^{\frac{5}{2}}dx$

• A. 19.360
• B. 16.594
• C. 10.129
• D. 11.817

Answer: (C)

Evaluate the following integral:
${\int }_1^2\text{\hspace{0.17em}⁣}\frac{5}{x^2}dx$

• A. $\frac{5}{3}$
• B. $\frac{5}{2}$
• C. $\frac{15}{2}$
• D. $5$

Answer: (B)

Fundamental Theorem of Calculus Part 2 Example 2

So in this problem, we're asked to evaluate the following definite integrals. So as you can see, we have a few examples here where what we can do is apply the fundamental theorem of calculus part two to solve these examples. Now let's go ahead and start with this first one. Here we're asked to find the integral from zero to five of this function right here. So notice we have kind of this long polynomial.

And what I can do is apply the fundamental theorem of calculus, which says I can take this and find the antiderivative of this function, then I can bound it from zero to five, and then I can go ahead and find my result. Now the first step I'm going to do for taking this antiderivative is recall we learned about the sum and difference rule, which is I can integrate each portion or take the antiderivative of each portion separately for each piece that is being added or subtracted. So starting with this x cubed using the power rule, we'll get x to the fourth power over four. Then I can use the power rule on three x squared, and that's gonna give me three x to the third power over three. Then I can use the power rule on negative six x here, which is gonna give me negative six x squared over two.

Then we can have plus, and then this two is gonna be just become two x. And then that's all going to be bounded from zero to five. So this is the result we get for our polynomial. Now what I can do from here is try to simplify things a little bit. So this x to the fourth over four that's as simplified as it's going to get.

Here I can cancel the three is giving me x cubed. They're going to have minus and then six divided by two that's gonna give us three. So we're gonna have three x squared plus two x all bounded from zero to five. Now what I need to do is take my high bound and plug it in, take my low bound and plug it in, and subtract the results. So we'll start with our high bound here.

So plugging in my high bound, what I'm gonna do is plug in five for x. So we're going to have f5=544+53-3 · 52+2 · 5. That's what we plug in here, and notice that the next thing I need to plug in is this zero. This zero is just gonna make everything here go to zero since there's an x in every single portion. So putting zero in is just gonna make that whole thing zero.

So all you need to do is evaluate this whole thing, which would this whole result right here, and this is going to give you 865 over four if you combine the fractions and reduce as much as you can. So this would be the solution for example a. So as you can see, this process really just turns into finding the antiderivative and plugging in the numbers. And after that, it's really just a lot of arithmetic that you have to do to get to your final result. But we can see here clearly that 865 over four is the solution to the problem.

So that's how we can go about solving this. So this is the solution for example a, and now let's move on to our next example which is example b. So example b gives us this scenario right here where we have this fraction that we're dealing with 13 over y cubed, and we're integrating that with respect to y from negative two to four. Now what I'm going to do is find the antiderivative of this function and then bound it from negative two to four according to the fundamental theorem of calculus part two. Now, before I go ahead and find this antiderivative, it's gonna be kinda difficult with what we have so far.

But I can go ahead and take that 13 and bring it to the outside of the integral. I can also take this y cubed and write it as y to the negative third power since it's in the denominator here. So rewriting this we'll have 13 on the outside of this integral from negative two to four, and then I'm going to have y to the negative third power, integrated with respect to y. Now from here what I can do is find the antiderivative of this function. Well, I can use the power rule where I add one to the exponent and then divide by that new exponent.

So doing this for the y to the negative third power, well, we'll still have this 13 on the outside. Then we're going to have, in this case, we'll have y to the negative two, all over negative two, and then that's going to be integrated and then bounded from negative two to four. So this is what we get. Now from here what I can do is try to write this in a more simple way. So if I write this out, we're going to have 13, and then we're going to write this as it's gonna be negative, and then we would have one over two y squared.

I just took this y to the negative two and brought it down to the denominator and then made this whole thing negative. And then this is all gonna be bounded from negative two to four. Now at this point, what I can do is multiply in this 13 here. So it's gonna give me negative 13 over two y squared. That's all gonna be bounded from negative two to four.

So I'm just going ahead and writing this as simplified as I can get it before I apply these bounds. So now that we have this, we can go ahead and apply the bounds into our function. So let's go ahead and do that. Now I'll write this down here. So what we're gonna have first is plugging in this four, we're going to have negative 13 over two times four squared.

And then what we're gonna do is subtract off, and we're going to have negative 13 again for this lower bound, we're going to have negative, and that's gonna be 13 over, and it's gonna be two times negative two squared in the denominator. So what you could do is cancel these negative signs, making it a positive. And then we have negative 13 over two times four to the second power, and then it's gonna be plus, and we'd have 13 over two times negative two squared. And doing this whole result here should give you 39 over 32 when you combine the fractions. So 39 over 32 would be the result for example b, and that's how we can solve this problem.

So this is the process when it comes to dealing with definite integrals. As you can see, it can sometimes be a tedious process, but to make sure that you know how to do this rule where you take the antiderivative and solve the integral, that's how you're going to find your answer. Now lastly, we move to example c. Now my first step is going to be integrating this with respect to theta. Now I can do that by starting with this three over pi, which is a constant.

If I integrate each term separately, as we can do with the addition and subtraction rules for integration, well, I can take the antiderivative here, which is gonna give me three over pi times theta, then I can integrate this minus cosine theta, and the integral of cosine theta is sine theta, so it's gonna be minus sine theta, and then this whole thing is going to be bounded from zero to pi over two. Now let's go ahead and see what I can do from here. Well what I can do is take this pi over two and plug it into my function for theta. So we're going to have this whole thing equal to and then we're going to have three over pi, and it's going to be multiplied by pi over two, and then we'll have minus and then we're going to have the sine of pi over two.

So that's going to be this whole result here, and then what we're going to do is take this entire thing and subtract off our lower bound plugged in.

And our lower bound plugged in, well that's going to give us then, we'll have three over pi times zero minus the sine of zero. And this whole thing is just ultimately gonna go to zero when we do this out. So we have this as our result from here. And if we go ahead and evaluate this whole thing, this is all just gonna go to zero. So all you need to really do is evaluate this left side here, which should all come out to just one-half.

So one-half is the solution to example c, and that's how you can go about solving this problem. So as you can see, dealing with these definite integrals can be a tedious process, but once you really get the hang of dealing with the fact that it's gonna be the definite integral of this, then applying the bounds, it should become pretty straightforward, and it should become pretty consistent that you're able to solve these types of problems. So I hope you found this video helpful, and let's move on.