THE FUNDAMENTAL THEOREMS
Page Walkthrough
1. The Integrals
2. The First Fundamental Theorem of Calculus
3. The Second Fundamental Theorem of Calculus
2. The First Fundamental Theorem of Calculus
3. The Second Fundamental Theorem of Calculus
The Integrals
All right. By now, you presumably have checked out everything about Riemann sums. It's time to move onto the formal part integration. First, familiarize yourself with the below symbol. It's an elongated 'S'. Why? S stands for sum. Recall the last 'Sum' you have studied.
THE DEFINITE INTEGRAL
Now, we shall move on to the Definite integral. This is just basically what we prepped you for in Riemann sums. The following is the definition:
So, the definite integral of f(x) from a to b is just its area under the curve.
If the curve is below the x axis, the area is negative. Note that up to now, there appears to be no relationship whatsoever between the definite and indefinite integral. How can this be? After all, they are both integrals. Soon, you shall see. The Indefinite Integral
First, we learn something called the Indefinite integral. It is noted by the sign above.
The definition is very simple: It's the opposite operation of the derivative. And what does this mean? That means: Differentiate F(x), you get its derivative g(x). Now, the indefinite integral of g(x) is just F(x)+C. C is a constant. As you probably has seen, easy as it seems, it involves many hidden catches You are trying to reverse a process. Now, it's the power of deduction, not induction at work. And deduction? I hate it. You will Hate it too. Soon Enough. Can't believe how Sherlock Holmes lived long enough to be made into a book... To start with, the definite integral is not a function. It's a set of functions. It should have the from of something of x plus some constant. Why? The constant turns zero when it's differentiated, thereby eliminating the difference between f(x)+c and f(x)+2c when they are integrated. 
FIRST FUNDAMENTAL THEOREM OF CALCULUS
For now, we have studied the definition of definite and indefinite integration. However, it would be a labor to compute Riemann sums every time we approach an integration problem.
Luckily for us, Newton and Leibniz, the Brit and the European (logical error here), discovered the relationship between integrals and derivatives 300 years ago. There finding can be summarized with the First Fundamental Theorem of Calculus. Though this is not as useful as the second one we are about to study, it serves as a step stone to the second. If F(x) satisfies this, then:
A Way to Think About This
If you have prior experience in statistics, this is going to be much easier for you.
Think of F(x) as a gigantic paintbrush. It is brushing the area beneath the curve of f(t). So, the paintbrush starts at a value a, and at F(x), paints to x. It is only logical to assume that the speed at which F(x) increases (it's derivative), is equal to the next area it's going to brush. Now, as in calculus, the idea of infinity kicks in here. What if the next area is arbitrarily thin? It's just the value of the function evaluated at t, the next x in line as the paintbrush moves along the x axis. Now, in statistics, you might call F(x) the cumulative density function, the CDF, while f(x) is the probability density function, the PDF. Makes sense now? Statisticians? Drawbacks
After studying the first fundamental theorem of calculus, we have one question. The question is how the hell do we apply it on all definite integrals that passes through our hands? The answer is we can't. The reason is quite simple you should be able to tell why just by looking at the equation: this shit is for those "concept" multiple choices. In a test, they might dig out f(x) and give you F(x) and ask for their relationship. Beyond that, the first fundamental theorem of calculus carries less weight in the real world than a chicken turned upside down. (Save for its bridge to the second fundamental theorem of calculus, which is the KICKER!)

THE SECOND FUNDAMENTAL THEOREM OF CALCULUS
This is perhaps the most important thing the Integral Sector of this website contains. So, be sure not to waste it.
Well, we just learned about the First Fundamental Theorem of Calculus, and I remembered kicking a statement up there saying that the Second is the more useful one. This theorem will literally solve all definite integrals. Provided that you can find the indefinite integral of the function.
The second fundamental theorem of calculus goes like this:
Well, we just learned about the First Fundamental Theorem of Calculus, and I remembered kicking a statement up there saying that the Second is the more useful one. This theorem will literally solve all definite integrals. Provided that you can find the indefinite integral of the function.
The second fundamental theorem of calculus goes like this:
Where F(x) is the antiderivative, or indefinite integral of f(x). The proof of this theorem really lies with the first one. You can try on your own. I'll try to put it up in documents for download. But now, let's focus on what this theorem does. After all, in tests and in practice, it's the APPLIANCES of theorems that come into use, not the PROOFS. It's a whole different story for people who strive to be mathematicians, but they probably aren't reading this in the first place, so......
What does this say? First, it does not put any restrictions on the function lying inside the integral signs, nor on its lower bound and upper bound, a and b. This means this identity applies to ALL definite integrals. Secondly, we see this is a simple, simple expression, and it agrees with institution: the constant terms of the antiderivatives cancel, making out a definite number for the definite integral. We can calculate the definite integral with a hand calculator.
Now, an example would be this: integrate y=x, 0<x<1. Of course, some folks would try to find the area using the area of triangles. And I admit for this problem it's simpler. But let's use the definite integral. So, the antiderivative of y=x is y=0.5x^2+C, while C is a constant with any real value. Now, plug in x=1 and x=0 and minus the two expressions. (Note that C conveniently cancel.) You get what you want. The answer should agree with your triangle area.
After you convince yourself of the immaculately power of this theorem, seal it in your mind forever, next to the Chain Rule, Rules of Differentiation, and the next day for the dentist.
What does this say? First, it does not put any restrictions on the function lying inside the integral signs, nor on its lower bound and upper bound, a and b. This means this identity applies to ALL definite integrals. Secondly, we see this is a simple, simple expression, and it agrees with institution: the constant terms of the antiderivatives cancel, making out a definite number for the definite integral. We can calculate the definite integral with a hand calculator.
Now, an example would be this: integrate y=x, 0<x<1. Of course, some folks would try to find the area using the area of triangles. And I admit for this problem it's simpler. But let's use the definite integral. So, the antiderivative of y=x is y=0.5x^2+C, while C is a constant with any real value. Now, plug in x=1 and x=0 and minus the two expressions. (Note that C conveniently cancel.) You get what you want. The answer should agree with your triangle area.
After you convince yourself of the immaculately power of this theorem, seal it in your mind forever, next to the Chain Rule, Rules of Differentiation, and the next day for the dentist.