How to solve definite integrals

28 Oct 2019 Leave a Comment

Indefinite integrals are used to find the formula for the area under a curve of function f(x), whereas definite integrals allow you to calculate the value of the area. It’s like the area of a rectangle, b*h: this is the formula, and in order to find the value you have to plug in the values of b and h. The same is for definite integrals. A definite integral is generally written like this:

where a and b are called respectively lower and upper bounds.

Also,

As you can see, if you are given the definite integral of a function, treat it as an indefinite indegral, and then substitute the x‘s with the value of b and then of a, like this:

Another example:

And now the rules:

Bound inversion rule

The bounds of integration can be inverted by inverting the sign of the integral.

Example:

For the sake of appearance, you want to have the bounds in order, so if you are using letters, say d and h, you will have d as lower bound and h as upper bound; if you are using numbers you want the lowest to be the lower bound and the greatest to be the upper bound. This means that if you have something like this

you may want to write it as follows:

However the result does not change, so you can do whatever you prefer.

Also, if you have something like this

it would be better to write it as

Simmetry rule

There are two main types of functions:

• odd functions: if the input of these functions is negative, the result will be negative; if the input is positive (with the same absolute value of the negative input) the output will be positive:

Examples of odd functions are sin(x), x³, -x, x etc… .

• even functions: regardless of the sign of the input, the result will always be positive:

Examples are x², |x|, 1-x², cos(x) etc… .

If the integral of an odd functions has opposite bounds, like -3 and 3, 7 and -7… there’s no need to calculate it because the result will always be 0.

This is because the graph of an odd function is simmetric on the x-axis, so half of the area is negative and the other half is positive. If you sum up the two you get 0.

This is the area under the function f(x)=x from -1 to 1; the blue area is negative, the red one is positive. As you can see the areas are the same, but since one is positive and the other is negative the total area is 0.

When the integral of an even function has opposite bounds the following happens:

If we look at the graph of an even function, like x², we will see that it is symmetric on the y-axis, so the two halves of the area are positive and have the same values, which means the total area is twice the area from 0 to the bound a.

The two regions are shaded in red, showing that they are positive.

Coincident bounds

The integral from a to a of any function f(x) will always be 0, since from a to a no area is described.

Sum rule for continuous bounds

And here’s the proof:

---

At the beginning I said that to solve definite integrals what you have to do is treat it as an indefinite integral and then substitute the x’s with the bounds, like in this example:

We simply applied the power rule and then plugged in the values of the bounds. When you perform a substitution though (u-substitution, trigonometric substitution) you need to change the bounds, since you will no longer be integrating with respect to the original variable. The bounds are the values of x (if you are integrating with respect to x) at position a and at position b. When you use u-substitution, you need to write the bounds as values of u. For example

This means that

We can use u-substitution or integration by parts, but the former is faster and its purpose is to show you how the bounds change;

u=x², therefore u1=x1² and u2=x2² and so u1=4 and u2=9:

You don’t have to write u=4 and u=9, it’s just to make it easier to understand. Also, as you can see, there’s no need to undo u-sub because it’s a definite integral. Another way to solve definite integrals with substitutions is to treat them as indefinite integrals, undo u-sub at the end and then plug in the original bounds. The method I showed you is cleaner though, so I don’t recommend using the other one. When using other techniques, like integration by parts, the bounds do not change because the variable of integration remains the same.

How to solve improper integrals

In improper integral is a definite integral with at least one bound equal to infinity:

Notice: the integral can also be from 4 to infinity, -infinity to 7 etc…, not just +/-infinity and zero.

Solving these kinds of integrals in not as straightforward though, since we need limits. Simply put, you substitute the bound equal to +/- infinity with a letter, usually a if it’s the lower bound and b in case it is the other. This way you will solve the integral as usual and at the end you will insert the letter in F(x). After this you will calculate the limit. To make you understand this better I will show you:

Now, what does this mean? When we have a limit, in theory we have to substitute the number a variable tends to into the functions, like here:

Easy, right? But it’s not always like this;

What is infinity – infinity? You might be tempted to say 0, but in order to calculate it, we need to apply a rule called L’Hôpital’s rule, but I will talk about it in another post.

For now, this is all you need to know! If you have any doubt, question or suggestions, leave a comment and I will be happy to reply! Stay tuned!

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Integration, Limits Calculus, Definite integrals, how to solve definite integrals