Written by Rebekah

**O****nce upon a time, someone came up to you and asked you to compute the area under a curve. You were told that if you accomplished this feat, you would be awarded $1 million dollars.****So you grabbed some markers and made some pictures.**

**The area you were looking for was between the points a and b. So naturally, this is what you drew.**

**Inside the area you drew rectangles (see (a)). Each rectangle had a width of**

**Δx (Δ means you’re dealing with little pieces), and a height of whatever f(x) is. This then means that the area of each rectangle is:**

**Area = Base x Height = f(x)**

**Δx**

**You then added the areas of each rectangle together to get the approximate area under the curve.**

**Using math notation:**

**Area under curve ≈ Σf(x)**

**Δx**

**But you weren’t satisfied with the results, so you decided to make the rectangles even thinner (see (b)). This way, you’d be even closer to the exact area under the curve.**

**But then you got an idea!**

**Why not make**

**Δx so small, that it practically reaches zero (see (c)). Of course, this would mean that you would have an almost infinite amount of rectangles. So you decided it was time to get some new notation. The small widths that are almost zero would now be called dx instead of Δx, and the sum of these pieces would now be written as:**

**The a and the b mean that the little rectangles are being summed from a to b.**

**Basically you decided to use**

**∫ (which looks like an “s” for ∫um) instead of Σ.**

**Your idea was fantastic and you were extremely excited. You even decided you would call this new way of calculating almost-infinite sums “integrating.”**

**But all of a sudden, you realized that there was one huge problem: how were you going to calculate this huge sum (i.e. integral)?!**

**So you cracked your fingers, grabbed some more markers, and made yet another drawing.**

**“Hmmm…” you wondered, “what if there exists a function that describes the area under the curve?” So naturally you decided to call this function A(x). The next logical step was to call a small increase in area**

**ΔA. Because ΔA was so close to the area of a small rectangle (like in your first picture) you immediately realized that**

**ΔA ≈ f(x) Δx**

**Then, you did the following…**

**ΔA ≈ f(x) Δx**

**ΔA/Δx ≈ f(x)**

**<---divided both sides by Δx**

**You then decided you’d let those little pieces get so close to zero, but never actually reach zero. So…**

**dA/dx = f(x)**

**And then it clicked. Your eyes opened wide and you gasped in astonishment: “dA/dx is the derivative of A(x)!” you exclaimed.**

**Now all you had to do was find a way to undo the derivative dA/dx, so that you would end up with the function A(x). Of course, this would mean you would have to treat f(x) like a derivative, so that you could do the same thing on the other side of the equation (remember, you must perform equal operations on both sides of an equation in order to maintain both sides equal).**

**You decided to call the act of undoing a derivative “finding the antiderivative.”**

**So…**

**Antiderivative of dA/dx = antiderivative of f(x)**

**This had tremendous implications. And you summarized your conclusions as follows:**

**∫f(x) dx =**

**Sum of All Rectangles (i.e.**

**Integral) =**

**Antiderivative of f(x)”**

**Now all that was left was figuring out how to calculate antiderivatives. So you started simple.**

**“The derivative of x² is 2x, so this must mean that the antiderivative of 2x is x²,” you reasoned. But then you realized something very important—the derivative of**

**x² + 1 was also 2x. In fact, the derivative of x² + 2 , x² + 3 , and even x² + 157 will always be just 2x.**

**So, you decided you would say that the antiderivative of 2x is simply x² + C (C for any constant).**

**In order to not get confused later, the antiderivative of f(x) would be denoted F(x) + C.**

**∫f(x)dx = F(x) + C**

**(a capital F implies antiderivative).**

**So based on this notation,**

**A(x) = F(x) + C**

**But what then was C in your problem?**

**That was simple.**

**When the function is equal to A(a), this means the area is still zero.**

**So… A(a) = 0 = F(a) + C**

**Solving for C we get that C = - F(a); thus, A(x) = F(x) + C becomes A(x) = F(x) -F(a).**

**Because in your problem you were looking for the area from a to b, this would translate to: total area under curve = A(b) = F(b) – F(a).**

**At last, you had finished! You jotted down your results, and presented them to the guy who had offered you $1 million dollars.**

**The result was as follows:**

**The man handed you a check, and asked “do you have a name for this thing you have invented and/or discovered?”**

**With a glimmer in your eyes, and a huge smile on your face you answered and said “yes, it’s called the ‘Fundamental Theorem of Calculus.’”**

**THE END**