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Area and the antiderivative

Let's return to our function $f(x) = x^2 + 1$, and our quest to find the area under the graph $y=f(x)$ between $x=0$ and $x=5$, i.e.,

$$\int_0^5 (x^2 + 1) \; dx.$$

As we've seen, we can estimate the area by many narrow rectangles, each over a subinterval of $[0,5]$, with width $\Delta x$ and height given by a value of $f(x)$. As we take the limit of more and more narrower and narrower rectangles, we get a better approximation to the area.

To calculate the exact area, we introduce an area function $A$. For any fixed number $c > 0$, let $A(c)$ be the area under the graph $y=f(x)$ between $x=0$ and $x=c$. So we can define the function $A(x)$ to be the area between 0 and $x$. Clearly $A(0) = 0$. We seek $A(5)$.

Detailed description of diagram

The fundamental idea is to consider the derivative of the area function $A$. What is $A'(x)$?

In fact, $A'(x) = f(x)$. We can see this geometrically. To differentiate $A(x)$ from first principles, we consider

$$\frac{A(x+h) - A(x)}{h}.$$

Now $A(x+h)$ is the area under the graph up to $x+h$, and $A(x)$ is the area under the graph up to $x$. Thus $A(x+h) - A(x)$ is the area under the graph between $x$ and $x+h$.

Detailed description of diagram

If you were to 'level off' the area under the graph between $x$ and $x+h$, you'd obtain a rectangle (as shown in the graph above) with width $h$ and height equal to the average value of $f$; dividing the area by $h$ gives the average value:

$$\begin{align*} \frac{A(x+h)-A(x)}{h} &= \frac{\text{ area of rectangle }}{\text{ width of rectangle }} \\ &= \text{ height of rectangle } \\ &= \text{ average value of \(f\) over the interval \([x, x+h]\).} \end{align*}$$

If $h$ becomes very small, then the interval $[x,x+h]$ approaches the single point $x$. And the average value of $f$ over this interval must approach $f(x)$. So we have

$$\lim_{h \to 0} \frac{A(x+h)-A(x)}{h} = f(x).$$

In other words,

$$A'(x) = f(x),$$

and $A(x)$ is an antiderivative of $f(x)$.

Returning again to our example $f(x) = x^2 + 1$, what then is the area function $A(x)$? Its derivative must be $f(x) = x^2 + 1$. We're looking for a function $A$ such that

$$A'(x) = x^2 + 1 \qquad \text{and} \qquad A(0)= 0.$$

Thinking back to derivatives, we can see that the antiderivative

$$A(x) = \frac{1}{3} x^3 + x$$

satisfies these criteria. Hence the total area is

$$A(5) = \frac{125}{3} + 5 = \frac{140}{3} = 46 \frac{2}{3},$$

confirming our earlier estimates. We can write this as

$$\int_0^5 f(x) \; dx = A(5) - A(0) = \frac{140}{3}.$$

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