Definite Integrals

Definite integrals integrate a curve between a range for \(x\) [\(a\), \(b\)].

These have notation \(\int_{a}^{b}f(x)dx \), with a lower limit \(a\) and upper limit \(b\). When we integrate between two values for \(x\), we are finding the sum of infinitely small slices of that function between the range. Hence, the definite integral is used to calculate the area of the curve.

Before that, here's how to actually calculate a definite integral:

1. Calculate the indefinite integral, e.g. \(f(x)\)
2. Wrap the integral with square brackets with the range in right-bottom/right-top corner, then
3. Calculate \(f(b)-f(a)\)

1) Find \(\int_{2}^{3} 2x dx\)

\(\begin{aligned} &= \left[x^2 + c\right]_{2}^{3} \\[5pt] &= (3^2 + c) - (2^2 + c) \\[5pt] &= 9 - 4 \\[5pt] &= 5 \end{aligned} \)

You can see that the constant of integration, \(c\), cancels out with definite integrals. Thus, we don't need to bother writing it at all.

2) Calculate \(\int_{6}^{9} 3x^2 + 6x + 2 dx \)

\(\begin{aligned} &= \left[x^3 + 3x^2 + 2x \right]_{6}^{9} \\[5pt] &= (729+3(18)+18)-(216+3(36)+12) \\[5pt] &= (729+243+18)-(216+108+12) \\[5pt] &= 990-336 \\[5pt] &= 654 \end{aligned} \)

The Area Problem

The problem at hand: how do we find the area under the curve of a function? Well although I just told you, just imgagine I didn't.

Let's look at this graph:

What we want to do, is calculate the area under the curve between \(x=1\) and \(x=3\):

We can use the sum of multiple rectangles to estimate the area, kinda like this:

As we can see, when width of the rectangles decreases, the approximation for the area becomes more and more accurate.

This is called an 'underestimate', as the rectangles are drawn under the graph.

This is what an 'overestimate' looks like:

Hopefully you can see that an even better estimation would be subtracting an underestimation from an overestimation with both the same number/width of rectangles. Either way, accuracy can be improved by decreasing the width of the rectangles.

Representing the area problem in mathmatical notation

We can clearly see that a limit is involved, as the width of the rectangles approach 0. We can also see that the rectangles are all added together:

\(\displaystyle \lim_{x \rightarrow 0}\sum_{x=a}^{x=b} f(x) \Delta x\)

This here means in english: the sum of infinitely small slice of a function between values for x. This is how we calculate the area.

However, this equation can be represented equally with an integral:

\(\displaystyle \int_{a}^{b} f(x)dx \)

Negative functions

Should the function dip below the y-axis and form negative values for f(x), this formula would produce an incorrect area, as although the integral would decrease, the true area is still increasing. To prevent this, we must subtract the negative integrals from the positive ones. This results in once negative integrals becoming positive, resulting in the true area:

This is fairly easy if a graph of the function is provided. If not you would find all the roots (when (\y = 0\)), and all stationary points, finally determine their nature (local max or min) to find what portions are negative.

1) Find the area of \(f(x) = x^2 -1\) between \(x = 0\) and \(x = 3\)

\(\begin{aligned} A &= - \int_{-1}^{2} x^2 -4 dx + \int_{2}^{3} x^2 - 4 dx \\[10pt] &= - \left[\frac{x^3}{3} -4x \right]_{-1}^{2} + \left[\frac{x^3}{3} -4x \right]_{2}^{3} \\[10pt] &= - \left[(\frac{8}{3} -8) -(-\frac{1}{3} +4)\right] + \left[(\frac{27}{3} -12)-(\frac{8}{3} -8)\right] \\[10pt] &= -(\frac{9}{3} -12)+(\frac{19}{3}-4) \\[10pt] &= (- \frac{19}{3} +12) +(\frac{19}{3}-4) \\[10pt] &= \frac{10}{3} +8 \\[10pt] &= \frac{10}{3} +\frac{24}{3} \\[15pt] &= \frac{34}{3} \ \text{square units} \end{aligned}\)

Area between curves

The area between curves can be calculated by subtracting the bottom function from the top function

2) Find the area between \(y = -x^2 + 5\) and \(y = \frac{x}{2}\) between \(x=0\) and \(x=2\)

\(\displaystyle \begin{aligned} A &= \int_{0}^{2} (-x^2 +5)-(\frac{x}{2})dx \\[5pt] &= \left[ -\frac{x^3}{3} -\frac{x^2}{4} + 5x \right]_{0}^{2} \\[5pt] &= (-\frac{8}{3} -\frac{4}{4} +10) -(0) \\[5pt] &= -\frac{8}{3} + 9 \\[5pt] &= -\frac{8}{3} + \frac{27}{3} \\[15pt] &= \frac{19}{3} \ \text{square units} \end{aligned} \)