What is the geometric interpretation of the fundamental theorem of calculus?

[member 428835]

hey pf!

i'm trying to get a geometric understanding of the fundamental theorem: $$\int_a{}^{b}f'(x)dx=f(b)-f(a)$$ basically, isn't the above just saying that if we add up a lot of slopes on a line at every point we will get the difference of the y values?

thanks! feel free to add more or correct me

 

[mfb]

That's one way to express the theorem in words, indeed.

 

[PeroK]

To help understand the fundamental theorem, first, re-arrange it a little:

$$f(b) = f(a) + \int_a{}^{b}f'(x)dx$$
Now, imagine going from a to b in a number of small steps: a to x1 to x2 ... to b:

$$f(x_1) \approx f(a) + f'(a)dx_1 \ (dx_1 = x_1 - a)$$
$$f(x_2) \approx f(x_1) + f'(x_1)dx_2 = f(a) + f'(a)dx_1 + f'(x_1)dx_2 \ (dx_2 = x_2 - x_1)$$

So that:
$$f(b) \approx f(a) + \sum f'(x_{i-1})dx_i$$

And, the Integral is the limit of this as the steps get smaller:
$$f(b) = f(a) + \int_a{}^{b}f'(x)dx$$

 

[member 428835]

thanks!

 

[FactChecker]

Not just slope. Slope times dx, which is dy. So it is adding up many dys to get from y(a) to y(b).