here stevey, here is a little exposé i wrote on calculus to help teach it to dumb-dumbs just like you
PART 1 DERIVATIVES
h is the distance on the x axis between two points on a curve (a curve in the y=x2 +mx+b format) which form a secant line (a line intersecting two points on the curve).
We can find the slope of a tangent line of any point on the curve by finding the derivative:
f'(a)=[f(a+h)-f(a)]/h, as h approaches 0

so for y=x2 -8x+9, the derivative will be:
= ([(a+h)2 -8(a+h)+9]-[a2 -8a+9])/h
= (a2 +2ah+h2 -8a-8h+9-a2 +8a-9)/h
= (2ah+h2 -8h)/h =(2a+h-8)
= 2a-8

the derivative of the curve y=x2 -8x+9 is 2x-8
This will tell you the slope of the tangent line for any point on that curve by plugging the x value of that point into the derivative. So, for the point (3,-6) on the curve y=x2 -8x+9, you just plug the x value (3) into the derivative (which is 2x-8), which gives you -2.
So the slope of the tangent line at (3,-6) is -2.

You now know derivatives, that's half of calculus.

PART 2: ANTI-DERIVATIVES AND INTEGRALS

to find the area under a curve in the xn +mx+b format, we first need to find what is called an anti-derivative; for simple xn +mx+b curves, finding the anti derivative is simple. First, throw out the constant, 'b', you don't need it. Now, what you're going to do, for each piece of the x2 +mx function, you simply need to add 1 to its exponent then divide that piece of the equation by whatever number the exponent now is.
the antiderivative of xn will be (xn+1)/(n+1)
so, for the curve y=x2, the antiderivative will be x3 /3.
To find the area underneath that curve between the lines x=1 and x=3, simply plug both of those x values into the anti-derivative, and subtract.
So, for the area between x=1 and x=3 under the curve y=x2, we plug 3 into the antiderivative (which is x3 /3) getting 9, then plug 1 into the antiderivative getting 1/3.
9 - 1/3rd is 8 and 2/3rds, so the area is 8.6666666666 squared

The reason you don't need the constant is because you can just find whatever remaining area you need to with your standard length*width formula for finding the area of rectangles.

Congratulations, you now know the second half of calculus. I might go further in depth and add a post for finding the area under sine waves and stuff if there's a demand for it.