Why does definite integral define an area?

Question

I'm kind of new to integrals. I know that $$\int_a^b f(x)\,dx=\int f(b)-\int f(a)$$

Using definite integrals, I can calculate area between the function and the $x$ axis between $x=a$ and $x=b$.

For example, we have a function $\alpha(x) = x^2$. Now, the area between $y=0$ and $y=x^2$ between $x=0$ and $x=5$ is: $$\int_0^5x^2\,dx=\int5^2\,dx-\int0^2\,dx=\frac{5^3}{3}-\frac{0^3}{3}=\frac{125}3=41\frac23$$ Why does this concept work?

Answer 1

The situation is really a bit backwards from the way that it is often taught in calculus. First of all, for a continuous function $f$, we can define

$$\int_a^b f(x) dx$$

without reference to an antiderivative. This just involves limits of Riemann sums. Here I think the connection to geometric area is clear (at least when $f$ is a nonnegative function). If you haven't seen Riemann sums, get antiderivatives out of your head for a bit, read about Riemann sums, and then come back here.

Antiderivatives come in with the fundamental theorem of calculus. There are really two fundamental theorems of calculus. One tells you how to differentiate integrals with respect to their limits:

$$\frac{d}{dx} \int_a^x f(y) dy = f(x).$$

Historically, this was really the big deal: it gave us a recipe for computing antiderivatives, which opened the door to solving differential equations.

The other tells you that you can compute integrals by computing antiderivatives:

$$\int_a^b f'(x) dx = f(b)-f(a).$$

This is the one you are using when you compute definite integrals by finding antiderivatives.

The idea of the first fundamental theorem is that $\int_x^{x+h} f(y) dy$ is approximately the area of the rectangle of height $f(x)$ and width $h$. The idea of the second fundamental theorem is that a Riemann sum for $\int_a^b f'(x) dx$ amounts to adding up the change along the tangent line between $a$ and $a+h$, then $a+h$ and $a+2h$, then ..., then between $b-h$ and $b$. This should be approximately the change in $f$ itself, since the tangent line provides a "good" approximation.

Answer 2

Imagine the area under a curve $f(t)$ going from some arbitrary point, to a point x. Let's define a function $A(x)$ to signify the area.

Given that, what does the below mean to you? $$\frac{A(x + h) - A(x)}{h}$$ Well, $A(x + h) - A(x)$ Represents the area under $f(t)$ in the interval $h$. When we divide that area by $h$, we get the average value of $f(t)$ over the interval $h$. So, the above represents the average value of $f(t)$ in $h$. Now, if $h$ is infinitely small: $$\lim_{h\rightarrow0} \frac{A(x + h) - A(x)}{h} = f(t) $$ Of course, that's the definition of the derivative. $$A'(x) = f(t) $$ Or, oppositely, showing that we can relate the area under a function to its anti-derivative.