How does finding the area relate to finding an antiderivative? Suppose I want to find the area of a quarter-circle using integration. Let the equation of the curve be
$x^2+y^2=1$ where $x\geq0$ and $y\geq0$. The area required is the purple region.

Suppose we break this quadrant into small rectangular pieces as shown:

The area of each rectangular strip as a function of $x$ will be:
$$A(x)=\sqrt{1-x^2} \cdot dr$$
If we sum the areas of the rectangles, we get the area under the curve. Usually, however, to compute
$$
\int_{0}^{1}\sqrt{1-x^2} \, dx \, ,
$$
I just find an antiderivative of $\sqrt{1-x^2}$. However, as a high-school student, I'm not sure how this relates to summing the areas of the rectangles.
 A: The idea is that an integral is a limit of approximations by rectangular strips. So if you split the interval $[0,1]$ into $n$ strips of width $n$, in the strip from say $x=(k-1)/n$ to $x=k/n$, the area under the curve within that strip is between $\frac 1n\sqrt{1-(\frac{k}{n})^2}$ and $\frac 1n\sqrt{1-(\frac{k-1}{n})^2}$. (In general, we have to look at the minimum and maximum values of the function in the strip, but in this case it's easier since the function is decreasing, so the maximum will always be the left end.)
Summing this over all strips gives you bounds $$\sum_{k=1}^n\frac1n\sqrt{1-\Big(\frac{k}{n}\Big)^2}\leq A\leq\sum_{k=1}^n\frac1n\sqrt{1-\Big(\frac{k-1}{n}\Big)^2}.$$
If we increase $n$, we can get better and better bounds. If both bounds approach the same value as $n$ goes to infinity, then we know that value must be equal to the area. (This equality of the bounds happens for the sort of functions you're likely to encounter, although not for every function; in particular it always happens if the function is continuous.)
If you'd started from a simpler function (e.g. the area under $y=x^2$) you could compute these sums exactly and deduce what the limits are. In this case it's rather harder to do that. Most of the time, to actually compute an integral, we don't use the definition but rather the useful fact that integrals are antiderivatives. The intuitive reason for this, writing $F(x)=\int_0^xf(x)\,dx$, is that the difference $F(1+1/n)-F(1)$ is the same as taking one extra strip of width $1/n$ when approximating the area, so $F(1+1/n)-F(1)\approx \frac 1n\times f(1)$ (assuming that $f(x)$ is continuous at $x=1$, so that the height of the strip is close to $f(1)$ throughout). This means if you try to approximate the gradient of $F(x)$ at $x=1$, you get $$F'(1)\approx \frac{F(1+1/n)-F(1)}{1+1/n-1}\approx f(1),$$ and letting $n$ tend to infinity this gives you exactly the right answer.
A: Call the sum over rectangles $F(x)$. Whenever you add a new rectangle to your sum, you're adding $f(x)\Delta x$ to what you have before. So you have
$$
F(x + \Delta x) = F(x) + f(x) \Delta x.
$$
Now rearrange to get
$$
f(x) = \frac{F(x + \Delta x) - F(x)}{\Delta x}.
$$
Since the integral is the limit of $\Delta x \rightarrow 0$, we have
$$
f(x) = \lim_{\Delta x \rightarrow 0}\frac{F(x + \Delta x) - F(x)}{\Delta x} = F'(x).
$$
It may help your intuition to think of it in terms of position and velocity. If my velocity is $v(t)$, then in the short time interval $dt$, I travel a distance $v(t) dt$. If I add these all up, I get that the distance I traveled between times $t_0$ and $t_1$ is $\int_{t_0}^{t_1}v(t)dt$. But, of course, the distance I traveled is also just my final position minus my initial position, so we must have
$$
x(t_1) - x(t_0) = \int_{t_0}^{t_1}v(t) dt.
$$
And since velocity is just the derivative of position, $x(t)$ must be an antiderivative of $v(t)$.
