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Evaluation Theorem

- Calculus
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One of the most interesting topics is History for sure. The History of Mathematics is no exception, this way you can learn how ancient Mathematicians developed the tools you are using this day. Speaking of which, learning how the problem of finding areas below a curve was solved gives you a great insight into what kind of problems are tackled by Integral Calculus.

Finding areas below a curve using the limit of Riemann's sum is very insightful, but why go through all this hassle every time you need to evaluate a definite integral? An important part of Mathematics is the development of tools that facilitate calculations. One of these tools is the Evaluation Theorem, an important part of the Fundamental Theorem of Calculus. Here you will learn how to use the Evaluation Theorem to find the value of definite integrals.

The fundamental Theorem of Calculus comes in two parts: one that relates integrals and derivatives, and one that states how to evaluate definite integrals. Here is a quick review.

The first part of the Fundamental Theorem of Calculus states that, for a continuous function \(f(x)\) on an interval \([a,b]\), the function defined by

\[ F(x) = \int_a^x f(t)\,\mathrm{d}t\]

satisfies

\[ F'(x)=f(x).\]

The second part of the Fundamental Theorem of Calculus states that for a continuous function \(f(x)\) on an interval \([a,b]\), the definite integral can be evaluated as

\[ \int_a^b f(x)\,\mathrm{d}x = F(b)-F(a),\]

where \(F(x)\) is an antiderivative of \(f(x)\).

The second part of The Fundamental Theorem of Calculus is also referred to as **the Evaluation Theorem. **This important theorem gives you a tool for evaluating definite integrals, rather than taking the limit of a Riemann sum.

The **Evaluation Theorem** is another way of referring to the second part of the Fundamental Theorem of Calculus. You can also find it as the **evaluation part** of the Fundamental Theorem of Calculus.

The Evaluation Theorem receives its name from the fact that it can be used to calculate definite integrals by evaluating an antiderivative two times.

It is possible that you find the Evaluation Theorem written as

\[ \int_a^b f(x)\,\mathrm{d}x= F(x) \Bigg|_{x=a}^{x=b}\]

The above notation means to evaluate the function at \(x=a\) and subtract that value from the function evaluated at \(x=b,\) that is

\[ F(x) \Bigg|_{\;{x=a}}^{\;{x=b}} \equiv F(b)-F(a).\]

The concept of a definite integral originated from the problem of finding the area below a curve. Here is a quick refresher.

Suppose you want to approximate the area below a curve \( f(x)\) on the interval \([a,b]\). The first step would be to divide the interval \([a,b]\) into \(N\) subintervals. Next, you need to associate a rectangle with each subinterval because the area of a rectangle is easy to find. Finally, you add up the area of all the rectangles, and the result is an approximation of the area.

By taking the limit as \(N\to\infty\) you get infinitely many thin rectangles, so you get a perfect match of the area.

This process is rather tedious, isn't it? This is where the Evaluation Theorem comes in handy!

Rather than dividing the interval and finding approximations, you can get the exact area below the curve by just finding an antiderivative and then evaluating it!

Here is an example

Use the evaluation theorem to find the value of

\[ \int_2^5 x^2 \, \mathrm{d}x.\]

**Solution:**

In order to use the Evaluation Theorem, you first need to find an antiderivative of the integrand, which you can achieve with the help of the Power Rule, that is

\[ \begin{align} F(x) &= \int x^2\,\mathrm{d}x \\ &= \frac{1}{3}x^3+C. \end{align}\]

Next, you need to evaluate the above antiderivative at the integration limits, so

\[ \begin{align} F(5) &= \frac{1}{3}(5)^3+C \\ &= \frac{125}{3} + C, \end{align} \]

and

\[ \begin{align} F(2) &= \frac{1}{3}(2)^3+C \\ &= \frac{8}{3} + C. \end{align} \]

Finally, subtract \(F(2) \) from \( F(5)\), this will give you

\[ \begin{align} \int_2^5 x^2 \, \mathrm{d}x &= F(5)-F(2) \\ &= \left(\frac{125}{3}+C \right) - \left(\frac{8}{3}+C \right) \\ &= \frac{117}{3}. \end{align}\]

Note that, no matter what value of \(C\) you choose, it will cancel out! Because of this, you do not need to add the integration constant of the antiderivative used in the evaluation theorem. How convenient!

You have seen that the Evaluation Theorem can be used to find the value of a definite integral by just evaluating an antiderivative at the integration limits of the integral. This means that the evaluation theorem can be applied whenever you need the value of a definite integral! Here are a few application examples.

The Net Change Theorem is a formula for obtaining the new value of a changing quantity. It considers the integral of the rate of change of a function

\[ F(b) = F(a) + \int_a^b F'(x)\,\mathrm{d}x.\]

The Net Change Theorem is used when you know a certain value of a changing quantity along with its rate of change. Note that the formula used for the Net Change Theorem can be obtained by isolating \( F(b) \) from the Evaluation Theorem!

Suppose you have a function \( f(x)\) that is integrable in the interval \( [a,b]\). An accumulation function is a function \( F(x) \) such that

\[ F(x)= \int_a^x f(t)\,\mathrm{d}t,\]

or

\[ F(x) = \int_x^b f(t)\,\mathrm{d}t,\]

where \(a<x<b\).

Basically, an accumulation function is a function obtained by solving a definite integral while leaving one of the integration limits as a variable.

The Evaluation Theorem can be proved using accumulation functions, as you will see further below.

When finding the average of a set of values you first need to add all the values together, and then you need to divide that sum by the number of values you added. However, this method is different when finding the average value of a function.

To start off, you need to know over which interval you are going to find this average. Next, rather than adding all the values, you find the **definite integral** of the function over the given interval. Finally, you need to divide by the length of the interval. This method can be summarized by using the formula

\[ f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,\mathrm{d}x.\]

You might already know that the integral of a function over a given interval gives you the area below its curve, but what if you need to find the area between two curves? You just subtract the area of the function that is closer to the \(x-\)axis!

Suppose that \( f(x) \) and \( g(x) \) are integrable on the interval \( [a,b]\) and that \( f(x) \geq g(x) \) on the same interval. Then the area between \( f(x) \) and \( g(x) \) is given by

\[ A = \int_a^b \left( f(x) - g(x) \right) \, \mathrm{d}x.\]

There are different ways of proving the Evaluation Theorem. Here you can see a proof that relies on the first part of the Fundamental Theorem of Calculus.

Begin by defining the accumulation function

\[ g(x)= \int_a^x f(t) \, \mathrm{d}t.\]

The first part of the Fundamental Theorem of Calculus tells you that

\[ g'(x)=f(x).\]

Now suppose you pick up an antiderivative of \( f(x) \), this means that

\[ F'(x) = f(x)\]

as well. In general, \( F(x) \) and \( g(x) \) will differ from one constant, that is

\[ F(x)= g(x)+C.\]

From here you can evaluate \( F(a) \) and \( F(b) \), so

\[ F(a) = g(a)+C\]

and

\[ F(b) = g(b) + C.\]

By taking the difference of the two above expressions, you will find that

\[ \begin{align} F(b)-F(a) &= (g(b)+C)-(g(a)+C) \\ &= g(b)-g(a) \\ &= \int_a^b f(t) \, \mathrm{d}t -\int_a^a f(t)\,\mathrm{d}t. \end{align} \]

For the second integral, note that its integration limits are the same, so that integral is equal to \(0\).

Finding the integral \( \int_a^a f(t) \, \mathrm{d}t\) is essentially finding the area below a point, which is just a line. Since lines are one-dimensional objects, they have no area, so the definite integral is equal to \(0\).

This way, you can simplify the above expression, so

\[ F(b)-F(a) = \int_a^b f(t)\,\mathrm{d}t.\]

By rewriting the above equation, and by switching the integration variable from \(t \) to \(x,\) you get the usual form of the Evaluation Theorem, that is

\[ \int_a^b f(x)\,\mathrm{d}x = F(b)-F(a).\]

As mentioned before, definite integrals are usually found with the help of the Evaluation Theorem. Here are more examples.

Find the value of the definite integral

\[ \int_{-3}^{-1} (x+3)^3\,\mathrm{d}x.\]

**Solution:**

Begin by finding the antiderivative of the function in the integrand. You can expand the binomial and then use the Power Rule, but this is not advised, as it will make a longer calculation. Instead, you can use Integration by Substitution by letting

\[ u=x+3,\]

so

\[ \mathrm{d}u=\mathrm{d}x.\]

Doing this integral by substitution will give you

\[ \begin{align} \int (x+3)^3 \, \mathrm{d}x &= \int u^3\,\mathrm{d}u \\ &= \frac{1}{4}u^4. \\ &= \frac{1}{4}(x+3)^4. \end{align}\]

Next, you can use the Evaluation Theorem to find the definite integral, so

\[ \begin{align} \int_{-3}^{-1} (x+3)^3 \, \mathrm{d}x &= \left(\frac{1}{4}(-1+3)^4 \right) - \left(\frac{1}{4}(-3+3)^4 \right) \\ &= \frac{1}{4}(2)^4-\frac{1}{4}(0)^4 \\ &= 4. \end{align} \]

Here is an example using trigonometric functions.

Find the value of the definite integral

\[ \int_0^\pi \cos{x} \, \mathrm{d}x. \]

**Solution:**

This time you can find the antiderivative of the cosine by looking at the derivatives of trigonometric functions. You will find that

\[ \int \cos{x}\,\mathrm{d}x = \sin{x}+C,\]

but as usual, you do not need the integration constant for a definite integral. This way you can evaluate the definite integral, so

\[ \int_0^\pi \cos{x}\,\mathrm{d}x = \sin{\pi}-\sin{0} \]

Since the sine of \(\pi\) and the sine of \( 0 \) are both equal to \(0\) the definite integral is \(0\), that is

\[ \begin{align} \int_0^\pi \cos{x}\,\mathrm{d}x &= 0-0 \\ &=0. \end{align} \]

Now take a look at the graph of the cosine function in the interval from \(0\) to \(\pi\).

You can see that half of the area is above the \(x-\)axis, and the other half is the same, but it is below the \(x-\)axis, so that portion is negative. Since a definite integral gives you the net signed area, it makes sense that this result is equal to \(0\).

Finally, you can look at an example that involves logarithmic functions.

Find the value of the definite integral

\[ \int_0^1 \frac{1}{x+1} \, \mathrm{d}x. \]

**Solution:**

Once again you need to find the above integral by using substitution. Begin by letting

\[ u=x+1\]

so

\[\mathrm{d}u=\mathrm{d}x.\]

This way the indefinite integral becomes

\[ \int \frac{1}{x+1}\,\mathrm{d}x = \int \frac{1}{u} \, \mathrm{d}u,\]

which is one of the Integrals Involving Logarithmic Functions, by looking at a table (and ommiting the integration constant) you will find that

\[ \int \frac{1}{u}\,\mathrm{d}u = \ln{u},\]

so

\[ \int \frac{1}{x+1}\,\mathrm{d}x = \ln{(x+1)}.\]

Finally, evaluate the antiderivative at both integration limits, that is

\[ \begin{align} \int_0^1 \frac{1}{x+1}\,\mathrm{d}x &= \ln{(1+1)}-\ln{(0+1)} \\ &= \ln{2}-\ln{1}. \end{align}\]

Since the natural logarithm of \(1\) is \(0\), this simplifies as

\[ \int_0^1 \frac{1}{x+1}\,\mathrm{d}x = \ln{2}.\]

- The Evaluation Theorem is another way of referring to the second part of the Fundamental Theorem of Calculus.
- The Evaluation Theorem receives its name from the fact that it is used to find the values of definite integrals by evaluating antiderivatives.
- The Evaluation Theorem comes in handy whenever you need to find the value of a definite integral. It is an efficient alternative to the evaluation of a definite integral through the limit of Riemann's sum.

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