# Intuitive explanation for why the definite integral gives the area between the function and the x-axis

Could somebody please give an intuitive explanation for why the antiderivative of a function evaluated at $b$ minus the antiderivative of the function evaluated at $a$, where $b>a$, gives the area between the function and the $x$-axis between these two $x$ values.

It does not make much sense to me, could somebody please give an intuitive proof or intuitive explanation

• see for example (drcruzan.com/FTOC.html) – Jean Marie Apr 20 '17 at 21:13
• Thankyou but for the proof of the first fundamental theorem of calculus, the area under the curve is equal to F(b) - F(a) whether or not the limit of delta x tending to 1 is evaluated, this seems a little bit odd to me because this would mean that the number of rectangles or little areas that are used in the summation is irrelevant. If 4 rectangles were added it would give the same sum as if infinite rectangles were added together. Am I going wrong somewhere, I have probably misunderstood the proof? – Nav Hari Apr 20 '17 at 21:39
• For another post, you might check out this thing that I wrote for my students once upon a time. The relevant bit to you is section 1.3, and especially the content after definition 4. – davidlowryduda Apr 20 '17 at 23:36
• Have you looked at any of the related questions in the list to the right? At least two of them directly address your question. – amd Apr 20 '17 at 23:49

Let's assume we have a velocity function $v(t)=10$. The integral of velocity is position. The integral is the sums of the y-values (velocity) at infinitesimally small $t$ intervals between $a$ and $b$.

Let's do an integral from $a=2$ to $b=4$:

$$\int_2^4{10\mathrm{d}t}\\ =x(t)\Big{|}_2^4\\ =5t\Big{|}_2^4\\ =(5\times4)-(5\times2)\\ =10$$

What does that 10 mean? It represents the change in position BETWEEN two limits $a$ and $b$. The reason that is important is that your definite integral with limits $a,b$ is essentially integral of v(t) from a to b equals antiderivative at b minus antiderivative at a.

Let $x=\int{v(t)}\mathrm{d}t=5t$ be our position function. If you evaluate your antiderivative at $b$, then you are integrating the entire domain of the function up to $b$... for our position and velocity functions, the lower end of the domain is the beginning of time!

When you subtract the antiderivative evaluated at $a$, you chop off everything that came before $a$, so you're only calculating how much your position changed over your particular interval.

Chop up the interval $[a,b]$ into tiny subintervals $[x_i,x_{i+1}]$. Clearly the total change $f(b) - f(a)$ is equal to the sum of all the little changes $f(x_{i+1}) - f(x_i)$. But, $f(x_{i+1}) - f(x_i) \approx f'(x_i) (x_{i+1} - x_i)$. Thus, $f(b) - f(a) \approx \sum_i f'(x_i)(x_{i+1} - x_i) \approx \int_a^b f'(x) \, dx$. When we chop up $[a,b]$ more and more finely, the approximations get better and better, so by a limiting argument we discover that $f(b) - f(a) = \int_a^b f'(x) \, dx$.

• From this the area would be equal to the definite integral whatever the subintervals are. the area should surely be different if the number of rectangles changes. Surely, the the area should only be equal to the definite integral when the number of rectangles tends to infinity and the width's of the rectangles tend to 0. – Nav Hari Apr 22 '17 at 20:35
• @NavHari But notice that I have only approximate equalities, not strict equalities, throughout the argument. When the subintervals are extremely tiny, it seems plausible to expect (or at least hope) that the approximations are very good. We have shown that $f(b) - f(a) \approx \int_a^b f'(x) \,dx$, but it seems plausible that the approximation can be made as close as we like by using extremely tiny subintervals. It follows from this "limiting argument" (which could be made rigorous, with additional work) that we actually have $f(b) - f(a) = \int_a^b f'(x) \, dx$ (with exact equality). – littleO Apr 22 '17 at 22:51
• But even if the subintervals are not of width dx, the area would still equal F(b) - F(a). This surely means that the area of each rectangle of width dx is equal to F'(c)dx, with no approximation regardless of the limiting argument. Is there any reason for why this must be true, as it does not seem very intuitive. – Nav Hari Apr 23 '17 at 17:22
• @NavHari Are you invoking the mean value theorem to say that there exists a number $c_i \in (x_i,x_{i+1})$ such that $f(x_{i+1}) - f(x_i) = f'(c_i)(x_{i+1} - x_i)$ with exact equality? You are correct that we can do this, and that is an important step towards making our intuitive argument into a rigorous proof. The standard proof of the fundamental theorem of calculus does exactly this. I don't find it to be unintuitive, though, because the mean value theorem is itself an intuitive theorem. – littleO Apr 24 '17 at 5:40