I figured out how integration gives the area under a graph

 I learned how differentiation works a few months ago. Then I learned integration too. I was taught that integration is the inverse of differentiation. Hmm... okay. Then I was taught that the area under the curve of a graph drawn by a function f(x) can be obtained by integrating f(x). But I could not understand how that works. So I searched the web but it didn't helped. I found many online content about the topic but they just explained the theory, not the concept behind the theory.

I tried to understand this my self. By the way, I'm just a student trying to share my knowledge. So please let me know, if I'm wrong .

Suppose there's a function called g(x). Following image shows 2 graphs, y = g(x) curve on the left. Curve in the  right side is a representation of gradient value for each point of y = g(x) against its respective x - coordinate. 

To plot the second graph, we need to calculate the gradient of each point of the first graph. If the function is linear, we could just calculate the gradient by considering 2 points on graph. We do the same thing even for this non-linear function. But since the gradient changes rapidly, the distance between the 2 points we select should be so small to be accurate. That's what we do in differentiation. I'm assuming you know this process.

Please assume that the red line segment on the curve in the left is a small tiny point on the curve. Assume that this point has the height (vertical thickness) Δy (the green line on y axis) and width (horizontal thickness) Δx. 

Let's calculate the gradient on red point (differentiation).  Gradient of the red point will be Δy/Δx.

Gradient = Δy / Δx

Next, we mark the calculated gradient value on the second graph. That point is marked in red on the second graph (the red line segment). It's vertical distance from origin (y value) is Δy/Δx, it's thickness is Δx. It's horizontal distance from origin is same as the red point on the first graph.

Now I want to calculate the area under the curve of the second graph i.e y = g'(x). To do that, I will calculate the area under the red color point. Red color point is just a tiny point on the curve. If we calculate the area under each point as we do for the red point, and sum them up, we will get the area under the curve for y = g'(x).

Area under the red point = (Δy/Δx) * Δx = Δx

Area under the red point looks like a trapezium but actually, the red point is just a tiny point. So, we are good to go by considering that area as a rectangle.

By summing up all the areas under each point we get the area under the curve for y = g'(x).

Let's see what actually happened. First, we get the gradient of the red point for y = g(x). We did that by dividing Δy by Δx. So that's Δy/Δx. Then we plotted  it on the second graph i.e y = g'(x). To get the area under the red point of second graph, we multiplied it's height by width. So we multiplied Δy/Δx by Δx. We first divided Δy by Δx to get the gradient and then multiplied it again by Δx. So to get the area, we did the reverse process of calculating the gradient.

In reality, red point is so small that Δx and Δy is zero or near zero. Hence, to calculate gradient, we need to differentiate. As I explained above, to calculate the area under red point for second curve, we reversed the action we did to calculate gradient i.e we did the inverse of differentiation i.e integration. 

That's how integration gives the area under a curve. If you read this (unlikely to happen because google has no reason to show you this article) please leave something in the comments!