Definite Integrals: the Geometric Meaning of Definite Integrals (Part II)

Definite Integrals (Part II)

Let’s take a look at further properties of definite integrals.

Some limits, such as lower and upper limits, can be exchanged. When they are exchanged, the value is just equal to the negative of the original one.

If the lower limit and upper limit of a definite integral are identical, then the definite integral is equal to zero.

In this case, it is clear that the region has zero width, and therefore the area must be equal to zero.

The third property states that if you multiply a function by a constant C and then integrate, the result is equal to the integral of that same function with respect to x multiplied by C.

The explanation is simple. The original function is blue, and if you multiply a constant C to f(x), it becomes green. Therefore, the area of the green function is C times that of the blue one. That's why we have this property of constant multiple.

Get quality help now

Prof. Finch

Verified writer

Proficient in: Math

4.7 (346)

“ This writer never make an mistake for me always deliver long before due date. Am telling you man this writer is absolutely the best. ”

+84 relevant experts are online

Hire writer

In addition, if we consider the definite integral of two functions f(x) and g(x), then the definite integral of the sum of f(x) and g(x) is equal to the definite integral of f(x) plus the definite integral of g(x).

We can see this easily from the following diagram. The light brown region is defined by the function f(x). The blue region is defined by the function g(x). If we add these two functions together, we get the pink region shown here.

So the sum of brown and blue regions is equal to pink region when f(x) = g(x) or f(x) -g(x).

Suppose we take the definite integral from a to c. We can split it up into two pieces, from a to b and then from b to c, because from a to c we can split into two pieces.

The min-max inequality is expressed as follows. The definite integral of a function f(x) from a to b is less than or equal to the maximum value of f(x), times b minus a.

Now, this is actually the area of this pink region. It is determined by finding the maximum of the function times the total width here, b minus a. This outer rectangle has an area of b minus a.

On the left side, the minimum of f(x) times b minus a is the area of this green triangle, which is below the blue one. Therefore the area of the blue one lies between the lower one and the upper bar. This is called max-min inequality.

We also have the property of domination. Imagine that f(x) exists and is always greater than or equal to g(x). In other words, the pink graph is always below the green one in this figure. However, g(x) is always less than or equal to f(x).

Then, if f(x) is continuous and differentiable on [a,b], then the definite integral of f(x) over [a,b] is greater than or equal to the definite integral of g(x) from a to b. This is clear because the area of the pink region is less than or equal to the area of the green region. We can actually give a new definition to the definite integral of a function f(x) over a and b.

If you take the definite integral of a function f(x) from a to b and divide by the total length of the interval, that means b minus a, this is defined to be the average value of f(x) over the interval [a, b].

The reason is as follows: Suppose we choose arbitrarily values, c1, c2, c3 up to cn. Then we can sample a number of points between a and b. If we take the average value of these sampled points, it means that we add up f(c1), f(c2), all the way to f(cn) and divide by n.

So this is the average value of these sampled values of the function. It is equal to one over n times the sum of f(x) at each of the ck, divided by n. We divide this interval into n equals sub intervals, where the width of each sub interval is equal to b minus a divided by n; we will call this delta x.

One over n equals delta x divided by b minus a.

We move the term delta x inside the summation symbol, which gives us the Riemann sum.

The Riemann sum of the function when n tends to infinity is therefore equal to the definite integral. From these equations, we see that why one over b minus a times a definite integral is defined as the average value of the function is because this is the average of the sample points of end points of the function.