[Calculus] What is Calculus?

And now it is time for the final Action Project for my calculus course! In this unit, we have been learning about integrals, which are the opposite of derivatives. We learned how to do the reverse power rule, in which you add one to the exponent and then divide by that number. The integral is used for finding the area of the space that fills under a line on a graph. For this Action Project, we have been assigned two numbers to create an equation from. We then must estimate the area underneath the equation using Riemann sums, which is essentially where we create shapes underneath the line to estimate the area. After that, we have to use integrals to calculate the exact area underneath the line. Here is my project!

For my project, I chose the equation 6x-9x^2. I chose this because it has x twice and that’s pretty nifty I would say. It looks like this:


For the Riemann sum, I used the right Riemann sum strategy where the rectangles are touching the curve on the top right corner. Using this, I estimated that the area is about 0.44. I did this by calculating the area of each rectangle. The equation for that was 0.35*0.1+0.75*0.1+0.96*0.1+1*0.1+0.84*0.1+0.5*0.1=0.44. It can also be converted into function notation to be more accurate: 0.1(f(1)+f(2)+f(3)+f(4)+f(5)+f(6)). It looks like this:




The next step is to find the integral by doing the reverse power rule. First, we add 1 to both exponents turning it into 6x^2-9x^3. Then we divide each term by the new exponent by making it 3x^2-3x^3. This is the integral. We can also double check this is right by using the normal power rule which is to multiply the coefficients by the new exponents and then subtract 1 from each exponent. First, we do the coefficients turning into 6x^2-9x^3by multiplying the 3s by the exponents. Then we do the exponents, subtracting 1 from each turning it into 3x-3x^2 which is what our original equation was. Using the integral, we can calculate the exact area by plugging in our numbers. We plug in 0.7 and 0 into the integral and subtract the latter giving us 3*0.72-3*0.73-0. This equates to about 0.441. The full notation looks like this:

∫00.7 6x-9x^2 dx

[3x^2-3x^3]00.7

3(0.7)^2-3(0.7)^3-3(0)^2-3(0)^3

1.47-0.648-0

0.441

This means that the area of the shaded area is 0.441 units2.





In conclusion, my estimate was very close to the exact area! I think the calculus course I have been taking has been very interesting and informative. Calculus is very useful when it comes to figuring out exact numbers that we would not have been able to otherwise. Using derivatives, we are able to figure out the slope of a curve, a form of line where the slope is constantly changing. Using integrals, we are able to calculate the area under a curve where there is no standard equation and would be incredibly difficult to calculate manually. I think that the purpose and meaning of calculus is to be as close to infinity as possible, as is the ultimate goal for humans. 


I thought this project was very interesting as well! It was fun to be able to create an equation ourselves and solve what we have been solving for other equations the entire course. Learning this portion of math helps fill in the gaps of questions I had before about infinity, curves, and a variety of other topics. If I were to do this project again, I might pick a more complicated equation but other than that, I think this was a really good project and course!