Definite integral

The definite integral is used to measure the area of shapes that are described by a function.

Motivation

Let’s take the function $f(x)=\sin (x)+\frac{x}{4}$. We might ask – how large is the area under the graph from the point a = 0 to the point b = 10? We are interested in the area bounded by the x-axis, the graph of the function f and the vertical lines at a and b. The diagram sums it up:

We are interested in the area highlighted in blue. How could we find it? We can split the whole blue area into several smaller areas, work out the area of each smaller piece and add all these partial areas together. To work out the area of a smaller piece with a formula, it must have a sensible shape. A rectangle is an obvious choice.

First we choose an increasing sequence of points x₀ to x_n such that x₀ = a, x_n = b. The larger n is, the more accurate our result will be. We choose the simple sequence x₀ = 0, x₁ = 1, x₂ = 2, …, x₁₀ = 10. From neighbouring points we form the intervals [x_{i − 1}, x_i], so we get the set of intervals

$$D=\{[0,1],[1,2],[2,3],\dots ,[9,10]\}$$

We call the first interval D₁, the second D₂ and so on. Next we find the minimum value of the function f in each interval D_i. For example, in D₁ the function f has its minimum value at x = 0, and f(0) = 0. In D₂ the lowest value is at x = 1 and $f(1)=\sin (1)+\frac{1}{4}\approx 1.0915$. And so on. We label these values. We denote the lowest value of the function in the interval D_i by m_i. So m₁ = 0, m₂ ≈ 1.0915 and so on.

Now we draw rectangles on the graph. The ith rectangle has width x_i − x_{i − 1} and height m_i:

We can see that each interval has one rectangle that lies entirely under the curve but is as tall as possible. The height of each rectangle in the ith interval is exactly m_i. If we now added up the areas of these rectangles, we would get an estimate of the area under the curve. The more rectangles we draw under the curve, the more accurate our estimate. We would work out the area of the ith rectangle as s_i = m_i · (x_i − x_{i − 1}), so the sum of all the rectangles would be:

$$s=\sum _{i=1}^{10}{m}_i\cdot (x_i-x_{i-1})$$

We can also go the other way: instead of taking the minimum value on the interval D_i, we take the greatest value. We label these values M_i. The rectangles would then look like this:

Again each interval has one rectangle, and it always has height M_i. The top edge of a rectangle touches the curve only at the local maxima of its interval. Adding up the areas of the individual rectangles gives us another estimate of the area under the curve. We would work out this area as

$$S=\sum _{i=1}^{10}{M}_i\cdot (x_i-x_{i-1})$$

Again, if we increase the number of rectangles, we get a more accurate estimate.

We now introduce the lower and upper sums. We say that s(D, f) is the lower sum and its value is

$$s(D,f)=\sum _{i=1}^n{m}_i\cdot (x_i-x_{i-1}),$$

where f is a continuous function on the interval [a, b], x₀ = a, x_n = b, D is a partition of the interval [a, b] into smaller intervals and n is the number of these intervals. We denote the upper sum by S(D, f) and it is equal to:

$$S(D,f)=\sum _{i=1}^n{M}_i\cdot (x_i-x_{i-1}),$$

where f is a function and D is a partition of the interval [a, b] into n intervals.

Put simply, the finer the partition D, the closer the values of s(D, f) and S(D, f) get, until in the limit they are equal. So there is exactly one real number A such that

$$s(D,f)\le A\le S(D,f)$$

for every possible partition D. What is A? The number A is exactly the area under the curve, because it is the only value that is greater than or equal to the total area of the rectangles under the curve and at the same time less than or equal to the total area of the rectangles “above the curve”.

Definition of the definite integral

From the previous section we know that for a continuous function f on the interval [a, b] we have s(D, f) ≤ A ≤ S(D, f) for every partition D. This number A is called the definite integral of the function f from a to b. We write

$$A={\int }_a^{b}f(x)\mathrm{d}x.$$

The fundamental theorem of calculus tells us that the following holds:

$${\int }_a^{b}f(x)\mathrm{d}x=F(b)-F(a),$$

where F is an antiderivative of the function f. The right-hand side is sometimes also written like this:

$${\int }_a^{b}f(x)\mathrm{d}x={[F(x)]}_a^b$$

The geometric meaning of the definite integral is then the area under the curve of a non-negative function f, as we have already seen. We started with the function $f(x)=\sin (x)+\frac{x}{4}$, so let’s work out the area on the interval [0, 10]. We work out the definite integral

$${\int }_0^{10}\sin (x)+\frac{x}{4}\mathrm{d}x$$

First we integrate the function f:

$$\begin{array}{rcl}\int \sin (x)+\frac{x}{4}\mathrm{d}x& =& \int \sin (x)\mathrm{d}x+\int \frac{x}{4}\mathrm{d}x\\ & =& -\cos (x)+c_1+\frac{x^2}{8}+c_2\\ & =& \frac{x^2}{8}-\cos (x)+c_1+c_2\end{array}$$

So the antiderivative F has the form $F(x)=\frac{x^2}{8}-\cos (x)+c$. We write out the definite integral:

$$\begin{array}{rcl}{\int }_0^{10}\sin (x)+\frac{x}{4}\mathrm{d}x& =& F(10)-F(0)\\ & =& \frac{{10}^2}{8}-\cos (10)+c-(\frac{0^2}{8}-\cos (0)+c)\\ & =& \frac{100}{8}-\cos (10)+\cos (0)+c-c\\ & \approx & 12.5+0.839+1\\ & \approx & 14.339\end{array}$$

The approximate area under the curve on the interval [0, 10] is 14.339.