Really couldn't catch the concept on epsilon and delta in limits

[azizlwl]

Really couldn't catch the concept on epsilon and delta in limits.

Let ∂x=x₂ - x₁
In finding a gradient the value ∂y is taken at certain value.
But in finding area using integral, the ∂y is seen to taken as zero. F(x₂)=F(x₁)

Maybe one multiplication and the other is division.

 

[Mark44]

Really couldn't catch the concept on epsilon and delta in limits.

Let ∂x=x₂ - x₁

I think you mean Δx = x₂ - x₁. Δx means "the change in x."
Δ - uppercase Greek letter delta
δ - lowercase Greek letter delta

This is like the difference between D and d in the Roman letters.

This symbol - ∂ - is used for partial derivatives.

In finding a gradient the value ∂y is taken at certain value.

You probably mean Δy, "the change in y," which would be f(x₂) - f(x₁).

The slope of a secant line between two points (x₁, f(x₁)) and (x₂, f(x₂)) on a curve is given by

$$m = \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$

The derivative (also called gradient) is the limit of either fraction above as Δx approaches 0, or as x₂ - x₁ approaches 0.

But in finding area using integral, the ∂y is seen to taken as zero. F(x₂)=F(x₁)

No, not really. The differential in an integral would be dx or dt (or other), not ∂x or ∂t. If you're asking questions about slope and deltas, you're probably getting ahead of yourself in asking about differentials.

Maybe one multiplication and the other is division.