# Definite Integration

## Introduction to Definite Integrals

Up to now, we’ve studied the Indefinite Integral, which is just the function that you get when you integrate another function.

The definite integral is actually a number that represents the area under the curve of that function (above the $x$-axis) from an “$x$” position to another “$x$” position; we learned how to get this area using Riemann Sums. (And don’t forget that with non-curved figures, we can get the area under a curve without using Calculus, but just using Geometry!)

It’s not obvious that an integral is an area under a curve, but it helps when you think of the equation $\text{Distance}=\text{Rate}\,(\text{Velocity})\times \text{Time}$: for the case when the $x$-axis of the curve represents time, and $y$-axis represents rate, the area (length times width) can represent a distance (or change in position). We can use this principle to determine how much something changes (for example, its distance) over time.

Here’s the formal Definite Integral as the Area of a Region:

## Properties of Definite Integrals

Definite Integrals have some properties; think of these properties just like the properties of any type of area. Most are somewhat obvious:

Now let’s do some problems that demonstrate the definite integral as an area:

Here are a few problems that illustrate the properties of definite integrals. Note that not all of these integrals may be areas, since some are negative; we’ll soon learn that if part of the function is under the $\boldsymbol{x}$-axis, the integral is a “negative area”.

## Fundamental Theorem of Calculus

Wow! This sounds important, doesn’t it? That’s because the Fundamental Theorem of Calculus is important; this theorem is used around the world every day to obtain areas (among other things) of all sort of objects. And the great thing about this theorem is it’s so simple to use, especially compared to some of the summing techniques we’ve used. This theorem is also called the “Net Change” Theorem. Here goes:

### Definite Integrals on the Graphing Calculator

You can evaluate definite integrals in the graphing calculator using the math fnInt(, much like you used the nDeriv( for derivatives.

Hit MATH and then scroll down to fnInt( (or hit 9). Put the lower and upper values for the interval and type in the function using the X,T,θ,n key, hitting the right arrow key in between each entry. Then put “$x$” after the “$d$” for “$dx$”, using right arrow key again. (In the case of $\displaystyle \int\limits_{1}^{2}{{{{x}^{4}}}}\,dx$, we had to use the arrow key twice, since we had to use it after the exponent to go back down.) You can also go back with the left arrow key if you need to make any changes. Then hit “Enter”:

Note that if you have graphed a function in Y=, you can also use 2nd trace (calc) 7 ($\displaystyle \int{{f\left( x \right)dx}}$) to integrate the function, and you will be asked to enter lower and upper limits.

Here are some  Definite Integration problems. Notice that when we are taking the definite integral of an Absolute Value function, we need to split the function at the points where the absolute values equals 0, and then, as we did in the Piecewise Functions section, either use the original function, or negate the function, depending on the sign of the function (without the absolute value) in that interval. Note that you can check these using fnInt (MATH 9) on your graphing calculator.

And, again, we can use the definite integral to get an area, if the $y$-values in the interval are greater than 0 (the function is completely above the $x$-axis). Note in the second problem, we have to solve for the $x$-intercepts, or zeros, and sketch a graph (or use a Sign Chart) to see where the function lies above the $x$-axis.

## Definite Integration and Area

You’ve probably realized by now (and I’ve hinted at it a few times) that to get the value of an integral under the $x$-axis, you can take the opposite (negative) of the area of that region. Thus, you have to really be careful if a problem calls for an integral of a region that is both above and below the $x$-axis, you have to basically add up the area above the $x$-axis and subtract the area below the $x$-axis.

But let’s think about it. If you were to take the absolute value of the function (so that everything moves above the $x$-axis), you would have the area! Here’s an example:

Here are a few more problems on Definite Integration and Area:

## Mean Value Theorem (MVT) for Integrals

We learned about the Mean Value Theorem for Derivatives here in the Extreme Value Theorem, Rolle’s Theorem, and Mean Value Theorem section. The Mean Value Theorem (MVT) for Integrals is a theorem that guarantees that a continuous function in an interval contains at least one point where that function is equal to the average value of the function.

What does this mean in plain English? All it really means is that for a continuous function between two different points, there is at least one point where the “$y$” value is equal to the average of the all the “$y$” values in that interval. In terms of geometry, this means that there exists a rectangle between the two points whose area is equal to the area under the curve of the function between those two points. Think of flattening a mountain so it fills the valleys perfectly.

Here’s the formal definition of the Mean Value Theorem for Integrals:

We’ll use this formula to solve problems where we find the “$c$” guaranteed by the Mean Value Theorem for an integral in a specific interval. We’ll also derive the Average Value of a Function from this formula (below). Here are the types of problems you might see for the Mean Value Theorem:

## Average Value of a Function

Now we can solve for the Average Value of a Function by dividing both sides of the Mean Value Theorem equation by $\left( {b-a} \right)$. The average value of a function is the $f\left( c \right)$ in the Mean Value Theorem equation.

Here are some Average Value Theorem problems:

Here is an application of the Average Value:

## 2nd Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus has to do with taking the derivative of the definite integral; we basically “undo” the integral to get the original function back. We have to be careful though, since we don’t always get the original function exactly as it was: it turns out (because of the Chain Rule), we have to multiply the function by the derivative of the upper limit of the interval. Also, to use this theorem, the lower bound must be a constant (although, if it’s not, it can be made into two integrals).

This theorem says that $\displaystyle \frac{d}{{dx}}\left( {\int\limits_{a}^{x}{{f\left( t \right)dt}}} \right)=f\left( x \right)$, or if $\displaystyle F\left( x \right)=\int\limits_{a}^{x}{{f\left( t \right)dt}}$, then ${F}’\left( x \right)=f\left( x \right)$.

It looks like the derivative of an integral (accumulation function) gets us back to the original integrand with just a change of variables. But let’s keep going, using the First Fundamental Theorem of Calculus:  $\displaystyle \begin{array}{l}\frac{d}{{dx}}\left[ {\int\limits_{a}^{{g\left( x \right)}}{{f\left( t \right)\,dt}}} \right]\\\,\,\,=\displaystyle \frac{d}{{dx}}\left[ {F\left( {g\left( x \right)} \right)-F\left( a \right)} \right]\\\,\,\,=\displaystyle \frac{d}{{dx}}F\left( {g\left( x \right)} \right)=f\left( {g\left( x \right)} \right)\cdot {g}’\left( x \right)\end{array}$, where $a$ is a constant. The $F\left( a \right)$ disappears, since the derivative of a constant is 0, but for the $F\left( x \right)$, we substitute the upper limit in for the variable, but then have to use the chain rule to multiply by the derivative of this function.

Here’s an example, where we solve the definite integral and then take the derivative back. Note that you won’t have to do this much work for each problem; we’ll see soon that we just substitute the upper limit in the integral function, and then multiply by the derivative of that upper limit.

Here’s the formal definition of the 2nd Fundamental Theorem of Calculus, with the basic instructions on how to solve these problems:

Here are some Second Theorem of Calculus problems:

## Using U-Substitution with Definite Integration

We learned how to use U-substitution (U-Sub) here in the U-Substitution Integration section, but let’s do some problems using Definite Integration. Remember this about U-sub Indefinite Integration:

With U-sub and Definite Integration, we can do these problems in one of two ways:

• Substitute the expression for “$u$” back and use the original values for the upper and lower bounds. Remember that the upper and lower bounds are in terms of “$x$”.
• Keep the “$u$” in the expression and solve for new upper and lower bounds (solve for “$x$” in terms of “$u$”). Then we don’t have to put the expression for “$u$” back in the problem! (We only want to do this if it’s straightforward to get “$x$” in terms of “$u$”.)

We’ll show both these methods; the main thing is to make sure your upper and lower bounds match the variable you’re plugging them in for:

Understand these problems, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Integration as Accumulated Change– you’re ready!

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