When we integrate a single function, what we are finding is the area between the curve and the axis, by slicing it up into an infinite number of very slim rectangles, and adding them all together. For example:
The area of a rectangle is \(Length \times Width\). In the case of the above graph:
So if the area of a rectangle is \(Length \times Width\), the area of one of these rectangles is \(\sqrt{x} \cdot dx\).
Then, we can add the area of all these rectangles together to find the total area under the curve.
The actual function of the integration is to add up all of these individual rectangles we talked about above, so that we can find the total area underneath the curve \(f(x)\) (i.e. between the curve and the x-axis):
\(Area = \int_a^b f(x) \ dx\)
The variables above and below the integration symbol, \(a\) and \(b\), are known as the bounds of the integration.
Example
The graph \(y=\sqrt{x}\) extends infinitely in the positive \(x\) direction, so without putting boundaries on the integration, the area would also be infinite. In the case of the above graph, the upper bound is \(x=1\), and the lower bound is \(x=0\). So:
\(Area=\int_0^1\sqrt{x} \ dx\)
This equation says that we want to find the area of all of the rectangles (or \(\sqrt{x} \cdot dx\)), between \(x=0\) and \(x=1\), and add them all up to find the total area between \(x=0\) and \(x=1\).
To determine this area, we perform the following calculations:
\(Area=\int_0^1 \sqrt{x} \ dx\)
\(=\int_0^1 x^{\frac{1}{2}} \ dx\)
\(=(\frac{2}{3}x^{\frac{3}{2}}) |_{x=0}^{x=1}\)
\(=\frac{2}{3}(1)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}\)
\(=\frac{2}{3}\)