The Fundamental Theorem of Calculus

Read this section to see the connection between derivatives and integrals. Work through practice problems 1-5.

Applications – The Future

Calculus is important for many reasons, but students are usually required to study calculus because it is needed for understanding concepts and doing applications in a variety of fields. The Fundamental Theorem of Calculus is very important to both pursuits.

Most applied problems in integral calculus require the following steps to get from the problem to a numerical answer:

In some cases, the path from the problem to the answer may be abbreviated, but the three steps are commonly used.

Step 1 is absolutely vital. If we can not translate the ideas of an applied problem into an area or a Riemann sum or a definite integral, then we can not use integral calculus to solve the problem. For a few types of applied problems, we will be able to go directly from the problem to an integral, but usually it will be easier to first break the problem into smaller pieces and to build a Riemann sum. Section 4.8 and all of chapter 5 will focus on translating different types of applied problems into Riemann sums and definite integrals. Computers and calculators are seldom of any help with Step 1.

Step 2 is usually easy. If we have a Riemann sum \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \Delta \mathrm{x}_{\mathrm{k}} on the interval [a,b], then the limit of the sum is simply the definite integral \int_{a}^{b} f(x) d x.

Step 3 can be handled in several ways.

If the function f is relatively simple, there are several ways to find an antiderivative of f (sections 4.6, parts of chapter 6 and others), and then Part 2 of the Fundamental Theorem of Calculus can be used to get a numerical answer.

If the function f is more complicated, then integral tables (section 4.8) or computers (symbolic manipulators such as Maple or Mathematica) can be used to find an antiderivative of  f. Then Part 2 of the Fundamental Theorem of Calculus can be used to get a numerical answer.

If an antiderivative of f cannot be found, approximate numerical answers for the definite integral can be found by various summation methods (section 4.9). These summation methods are typically done on computers, and program listings are included in an Appendix.

Usually the difficulties in solving an applied problem come with the 1st and 3rd steps, and the most time will be spent working with them. There are techniques and details to master and understand, but it is also important to keep in mind where these techniques and details fit into the bigger picture.

The next Example illustrates these steps for the problem of finding a volume of a solid. Problems of finding volumes of solids will be examined in more detail in Section 5.1.

Example 5: Find the volume of the solid in Fig. 7 for 0 \leq x \leq 2. (Each perpendicular "slice" through the solid is a square.)


Step 1: Going from the figure to a Riemann sum. If we break the solid into n "slices" with cuts perpendicular to the x–axis, at x_{1}, x_{2}, x_{3}, \ldots, x_{n-1} (like cutting a loaf of bread), then the volume of the original solid is the sum of the volumes of the "slices" (Fig. 8):

Total Volume = \sum_{\mathrm{i}=1}^{\mathrm{n}} (volume of the ith slice ).

The volume of the ith slice is approximately equal to the volume of a box:

    &\text { (height of the slice) } \cdot \text { (base of the slice)\cdot(thickness) } \\
    &\approx\left(\mathrm{c}_{\mathrm{i}}+1\right) \cdot\left(\mathrm{c}_{\mathrm{i}}+1\right) \cdot \Delta \mathrm{x}_{\mathrm{i}}

where \mathrm{c}_{\mathrm{i}} is any value between \mathrm{x}_{\mathrm{i}-1} and \mathrm{x}_{\mathrm{i}}.

Therefore, \text { Total Volume } \approx \sum_{i=1}^{n}\left(c_{i}+1\right) \cdot\left(c_{i}+1\right) \cdot \Delta x_{i} which is a Riemann sum.

Step 2: Going from the Riemann sum to a definite integral.

The Riemann sum approximation of the total volume in Step 1 is improved by taking thinner slices (making all of the \Delta \mathrm{x}_{\mathrm{i}} small), and

Total volume = \lim _{\text {mesh } \rightarrow 0}\left\{\sum_{i=1}^{N}\left(c_{i}+1\right) \cdot\left(c_{i}+1\right) \cdot \Delta \mathrm{x}_{\mathrm{i}}\right\}

=\int_{0}^{2}(x+1)(x+1) \mathrm{dx}=\int_{0}^{2}\left(x^{2}+2 x+1\right) \mathrm{dx}.

Step 3: Going from the definite integral to a numerical answer.

We can use Part 2 of the Fundamental Theorem of Calculus to evaluate the integral.

F(x)=\frac{1}{3} x^{3}+x^{2}+x is an antiderivative of x^{2}+2 x+1 (check by differentiating \mathrm{F}(\mathrm{x}) ), so

    &\int_{0}^{2}\left(x^{2}+2 x+1\right) \mathrm{dx}=\mathrm{F}(2)-\mathrm{F}(0)=\left\{\frac{1}{3} 2^{3}+2^{2}+2\right\}-\left\{\frac{1}{3} 0^{3}+0^{2}+0\right\} \\
    &=\left\{\frac{26}{3}\right\}-\{0\}=\frac{26}{3}=8 \frac{2}{3}

The volume of the solid shape in Fig. 7 is exactly 8 \frac{2}{3} cubic inches.

Practice 4: Find the volume of the solid shape in Fig. 9 for 0 \leq x \leq 2. (Each "slice" through the solid perpendicular to the x–axis is a square.)