Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.

Let's denote the width of a sample by $h$ where $$h\rightarrow0$$enter image description here

Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can add the areas of the rectangles like $$f(a)h+f(a+h)h+f(a+2h)h$$

but the definite integral $$\int_a^b f(x)h$$ suggests that we do $$f(a)h+f(a+h)h+f(a+2h)h+f(b)h$$

I'm not able to understand where I've gone wrong.

Please tell me whether my thinking is correct or not.

Why does thinking in these two different ways give us different results?

I have another question.

This is very closely related to the one I've asked above. So I decided not to ask a separate question.

In the property of definite integrals, where $$\int_0^{2a}f(x)dx=\int_0^a f(x)dx+\int_0^a f(2a-x)dx$$ the area of the sample $f(a)dx$ is counted twice. Why is it so? Doesn't it cause any change to the total area under the curve?


I'm not quite sure what you mean by "the definite integral suggests using $\int_a^b f(x)h$"

First of all, this is not true: the definite integral is $\int_a^b f(x)dx$. Assuming that the top of those boxes are $f(x)$, it would be evaluated precisely as $f(a)h+f(a+h)h+f(a+2h)h$. I don't know where you're getting $f(a)h+f(a+h)h+f(a+2h)h+f(b)h$.

I think the problem you're having here is that you're not making a necessary distinction between a definite integral and a Riemann sum, and, more importantly, that you aren't considering which type of Riemann sum you want. There are left, right, and middle Riemann sums, and which one you choose will determine how you represent those boxes. It seems like what you're trying to do with $f(a)h+f(a+h)h+f(a+2h)h+f(b)h$ may be to do both a left and right Riemann sum at the same time, and that doesn't work. You should read up on the different types of Riemann sums and compute them yourself; this should clear up the conceptual issue you're having.

Regarding your second question, it doesn't matter because the set $\{a\}$ has measure zero. The technical meaning of that is somewhat complicated, but essentially you can think of it as meaning that the "slice" of the integral at $x=a$ has height $f(a)$, but width $0$, and therefore has area $0*f(a)$.

Again, you should look at the Riemann sum definition of the definite integral here to fully understand what's going on. $\int_a^b f$ is a limit of these boxes, and viewing it as a limit is really the only way to resolve the intuitive flaws that come out of viewing the integral as adding up zoomed in boxes.

| cite | improve this answer | |
  • $\begingroup$ I got the first first part. But for the second part, I don't get what you mean by "set {a} has measure zero". Please help. $\endgroup$ – user8718165 Apr 27 '19 at 5:19
  • 1
    $\begingroup$ @user8718165 Like I said, it's a technical term, but the intuition behind it is that the overlap between the interval $[0,a]$ and the interval $[a,2a]$ consists of only the number $a$, which has zero width. $\endgroup$ – Alexander Gruber Apr 27 '19 at 5:22

The problem here is that $f(a)h$ and the other addends in the sum don't actually correspond to rectangles in your picture. You are actually trying to add $f(a+\frac{h}{2})h + f(a+\frac{3h}{2})h + f(a+\frac{5h}{2})h$.

Those are the rectangles depicted in your image.

And for your second question, in integrals $f(a)\,dx$ is infinitesimal and equal to $0$. Only when you put all the $f(x)\,dx together that you get a finite quantity.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.