# Trouble proving that the boundaries do not matter when finding the area under the curve of $f(x)$

I was writing a note to myself about the Fundamental Theorem of Calculus, but then I stumbled on the final hurdle. While I could prove that $$A(x)$$ tells you the area under the curve between $$x=0$$ and $$x=x$$, I could not generalise this proof to show that the definite integral between $$x=a$$ and $$x=b$$ tells you the area under any curve, no matter what the boundaries are.

In other words, I could prove that: $$A(x) = \int f(x)dx$$ But not: $$\text{Area}=\int_a^b f(x)dx$$

Here is what I had written:

The fundamental theorem of calculus links the concepts of derivatives and integrals together. While at first glance, finding the area under a curve seems unrelated to finding a gradient function, these represent 'opposite' operations. In other words, when we differentiate $$f(x)$$ to find $$f'(x)$$, we are computing the gradient function of $$f(x)$$; and when we integrate $$f'(x)$$, we can use $$f(x)$$ to find the area under the curve of $$f'(x)$$. Just as differentiation and integration are inverse operations, so too are finding the gradient function and finding the area under the curve.

The geometric interpretation of this theorem and its proof are as follows:

$$y=f(x)\\ \text{let A(x)= the area under the curve between x=0 and x=x}\\$$

The area between $$A(x)$$ and $$A(x+h)$$, where $$h$$ is a small number, is equal to $$A(x+h)-A(x)$$

$$A(x+h)-A(x)$$ is approximately equal to rectangle. This approximation ignores changes in the height of the curve, but as $$h$$ tends towards $$0$$, the error caused by this approximation also tends towards $$0$$. Therefore, we can say that:

$$A(x+h)-A(x)\approx \text{base \times height} = hf(x)$$

Dividing both sides by h gives: $$f(x) \approx \frac{A(x+h)-A(x)}{h} \\ f(x) = \lim_\limits{h \to 0} \frac{A(x+h)-A(x)}{h}$$

We also know the following is true from the First Principle of Differentiation:

$$A'(x)= \lim\limits_{h \to 0} \frac{A(x+h)-A(x)}{h}$$

Consequently,

$$A'(x) = f(x)$$

And integrating both sides gives us our final answer:

$$A(x) = \int f(x)dx$$

• OK, so what is $A(b)-A(a)$? Jan 14, 2020 at 18:13
• @Vasya $A(b) - A(a)$ is the area under the curve between $x=a$ and $x=b$! This can be represented as a definite integral because $\int_a^b f(x) dx$ means "integrate the function and find A(b) - A(a)". Thank you.
– Joe
Jan 14, 2020 at 18:43
• In other words, I could write: $\text{area under the curve between$x=a$and$x=b$} = A(b) - A(a) = \int f(b)dx - \int f(a)dx= \int_a^b f(x)dx$
– Joe
Jan 14, 2020 at 18:49

I usually "prove" this - admittedly with a bit of hand-waving - as follows: we write the theorem in the form $$\int_a^b f'(x)\,dx = f(b) - f(a)$$ and choose some segmentation of $$a=a_0 < a_1 < ... < a_n = b$$ of the interval $$[a,b]$$. Then the integral can be written (without limits) as the Riemann sum $$\sum_{i=0}^{n-1} f'(x_i)(x_{i+1} - x_i)$$ According to the definition of the derivative we might write $$f'(x_i)\approx\frac{f(x_{i+1})-f(x_i)}{x_{i+1}-x_i}.$$ Replacing $$f'$$ within the Riemann sum of the integral, we just get a telescoping sum, and the result follows. This reasoning is not completely rigorous, but might possibly be helpful anyways.