From the previous topic, we learned the area between a function and the \(x\)-axis is the limit of a Riemann sum:$$\lim_{\Delta\rightarrow0}\sum_{k=1}^{n}f(x_k^*)\Delta x_k$$However, this notation is difficult to write and manipulate. The definite integral provides a much cleaner way to express this area:$$\int_{a}^{b}f(x)dx=\lim_{\Delta\rightarrow0}\sum_{k=1}^{n}f(x_k^*)\Delta x_k$$The constants \(a\) and \(b\) are the limits of integration, \(f(x)\) is the integrand function and the \(dx\) is decribing the variable of integration. As a quick illustration, the velocity function from the Approximating Areas topic would be written as:$$\int_{0}^{8}t^2dt=170.\overline{6}$$
Consider the following graph:
This area considered for this function, has parts below the \(x\)-axis. This makes the values of \(f(x_k^*)\) negative in the Riemann sum.
How can area be negative? Nothing changes as far as the calculation goes, it is just interpreted as having area below the \(x\)-axis.
In this graph, the orange area is negative, but the green areas are positive, and the net area is the sum of these three areas. A net area can be positive, negative or zero.
Find the value of \(\int_{1}^{6}\left(2x-6\right)dx\) using Geometric area formulas.
Since definite integrals are limits, many properties follow through. Let \(f\) and \(g\) be integrable functions on \([a,b]\), \(p\) any value in \((a,b)\) and \(c\) any non-zero constant:
$$\int_{a}^{a}f(x)dx=0$$
$$\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx$$
$$\int_{a}^{b}\left(f(x)-g(x)\right)dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$$
$$\int_{a}^{b}cf(x)dx=c\int_{a}^{b}f(x)dx$$
$$\int_{a}^{b}f(x)dx=\int_{a}^{p}f(x)dx+\int_{p}^{b}f(x)dx$$
$$\int_{a}^{b}|f(x)|dx$$Turns negative areas positive.
Assume \(\int_{0}^{7}g(x)dx=-10\) and \(\int_{0}^{5}g(x)dx=3\). Use the properties of definite integrals to find each of the following: