My son is beginning his final year of secondary school this year and one of the major components is calculus. He's always been better at visual methods and I'm struggling to remember how I learnt this stuff some forty years ago.
My (vague) understanding is that differential calculus has to do with the slope of an equation at a point and this is achieved by choosing two coordinates (say at x=4 and x=5 when you're wanting the slope at the former) and working out the slope of the line between them. This gives an approximation of the slope which is perfect for a linear equation but not so good for any other equation.
Then, as we slide the second point closer to 4 (say using 4 and 4.5), the approximation tends to come closer to the actual slope (wildly curvy lines like those with terms up to x99 may not do this immediately but they'll get there eventually).
Hence we're looking for the limit of dy/dx as dx -> 0.
Similarly, integral calculus is area under the line between two points, down to the x-axis. The idea there is to form a shape under the line (triangle on top of rectangle, which I'm reliably informed in the comments by the good folk here that it's a right trapezoid).
The area of that shape is an approximation since its top is a single straight line which may not match the actual shape of the equation. Just like in the differential calculus described above, that top line of the right trapezoid will approach the actual equation as the two x co-ordinates get closer.
So, in this case, we still use limits but the idea is to gradually increase the number of right trapezoids in the region. Hence the first step would be to add the areas of the two between x = 4 to 4.5 and x = 4.5 to 5. This would make the two top lines a better approximation of the actual equation line, hence make the resultant area calculation closer (again, restricted at first by any wild gyrations in the actual equation).
Then the sum of the area of the four right trapezoids bounded by x = 4 to 4.25, x = 4.25 to 4.5, x = 4.5 to 4.75, and x = 4.75 to 5. Hence this is the sum of all areas as the number of right trapezoids approaches infinity (or their width approaches zero, if you'd prefer).
Is that a correct understanding of the two types of calculus, in terms of visualisation? Can anyone point out any obvious (or not so obvious) flaws in my reasoning?
As I said, it's been a while since I've done this stuff. I was quite good with all sorts of math concepts even through Uni but lack of practice has dimmed the memories somewhat. He's doing Comp Sci which is handy given my background, but I'd like to be able to help out with other subjects as well, just not that 🙂
Best Answer
You are correct. Just some further explanation:
Regarding the derivative part, here is a more precise notation: we usually say the slope is $\frac{\Delta{y}}{\Delta{x}}$ as $\Delta{x} \to 0$. And, we usually use the notation $\frac{dy}{dx}$ to show the exact derivative.
Regarding the integral, your explanation and approach is again correct. "Riemann sum" is also another kind of approximation for calculating the area under a curve and I think it might be easier due to the fact that it uses rectangular to the best of my knowledge. And I think rectangular is an easier shape.
But, overall, you are good to go.