1 Calculating the area under a curve

Let us denote the area under $y=f\left(x\right)$ between a fixed point $a$ and a variable point $x$ by $A\left(x\right)$ :

Figure 6

$A\left(x\right)$ is clearly a function of $x$ since as the upper limit changes so does the area. How does the area change if we change the upper limit by a very small amount $\delta x$ ? See Figure 7 below.

Figure 7

To a good approximation the change in the area is:

$A\left(x+\delta x\right)-A\left(x\right)\approx f\left(x\right)\delta x$

[This is because the shaded area is approximately a rectangle with base $\delta x$ and height $f\left(x\right)$ .] This approximation gets better and better as $\delta x$ gets smaller and smaller. Rearranging gives:

$f\left(x\right)\approx \frac{A\left(x+\delta x\right)-A\left(x\right)}{\delta x}$

Clearly, in the limit as $\delta x\to 0$ we have

$f\left(x\right)=\underset{\delta x\to 0}{lim}\frac{A\left(x+\delta x\right)-A\left(x\right)}{\delta x}$

But this limit on the right-hand side is the derivative of $A\left(x\right)$ with respect to $x$ so

$f\left(x\right)=\frac{dA\left(x\right)}{dx}$

Thus $A\left(x\right)$ is an indefinite integral of $f\left(x\right)$ and we can therefore write:

$A\left(x\right)=\int \; <!--nolimits\; -->f\left(x\right)dx$

Now the area under the curve from $a$ to $b$ is clearly $A\left(b\right)-A\left(a\right)$ . But remembering our shorthand notation for this difference, introduced in the last Section we have, finally

$A\left(b\right)-A\left(a\right)\equiv {\left[A\left(x\right)\right]}_a^b={\int \; <!--nolimits\; -->}_a^{b}f\left(x\right)dx$

We conclude that the area under the curve $y=f\left(x\right)$ from $a$ to $b$ is given by the definite integral of $f\left(x\right)$ from $a$ to $b$ .