How Do You Find the Area Between Curves?

In Section 14.3, you will learn how to find the area between two curves. Suppose you have two functions f(x) and g(x). Also assume that the higher curve is f(x). We are interested in finding the area from a point x = a to x = b between the two curves. We can do this by finding the area below f(x) and above the x-axis,

$latex displaystyle intlimits_{a}^{b}{f(x),dx}$

and subtracting the area below g(x) and above the x-axis,

$latex displaystyle intlimits_{a}^{b}{g(x),dx}$

Alternately, we can subtract the functions first and then find the area,

$latex displaystyle intlimits_{a}^{b}{left[ f(x)-g(x) right],dx}$

On Monday, the face-to-face class worked several of these types of problems.

Continue reading “How Do You Find the Area Between Curves?”

What is Integration by Parts?

This process for reversing the Product Rule for Derivatives is called Integration by Parts . It is covered in Section 14.2. In Integration by Parts, the integrand (the thing you are finding the antiderivative of) is written as a product. One piece is thought of as u and the other part v‘. The formula then says

$latex \int{u{v}’ dx=uv-\int{v{u}’ dx}}$

Below are several examples that students worked out.

Problem 1

$latex \displaystyle \int{\left( 1-x \right){{e}^{x}} dx}$

Problem 2

$latex \displaystyle \int {\left( 8x+10 \right) \ln \left( x \right) dx}$

Problem 3

$latex \displaystyle \int{\left( 2t-1 \right) \ln \left( t \right) dt}$


How Do You Find an Antiderivative Using the Substitution Method?

In Monday’s class, student found many antiderivatives using the Substitution Method. The basic process is illustrated below for the antiderivative

$latex \displaystyle \int{4{{\left( {{x}^{2}}-3 \right)}^{3}}\cdot 2x,dx}$.

Let’s look at the basic steps.

  1. Choose the expression for u. This is generally the inside part of a composition in the integrand. Use the derivative to find an other expression for du.
  2. Match the integrand with u and du. All variables in the original integrand must change to u.
  3. Change the integrand so that it is written in terms of u.
  4. Work out the antiderivative in terms of u.
  5. Put in the expression for u so that the antiderivative is written in terms of the original variable.

Now let’s look at the examples carried out in class.

Problem 1

$latex \displaystyle \int{{{\left( 3{{x}^{2}}+4 \right)}^{4}}\cdot 6x,dx}$

Problem 2

$latex \displaystyle \int{{{\left( 3{{x}^{2}}+4 \right)}^{3}}\cdot 4x,dx}$

Problem 3

$latex \displaystyle \int{{{\left( {{x}^{2}}-1 \right)}^{5}}\cdot x,dx}$

Problem 4

$latex \displaystyle \int{4{{\left( 2x+3 \right)}^{2}},dx}$

Problem 5

$latex \displaystyle \int \frac{2}{{{\left( 2m+1 \right)}^{3}}} dm$


Problem 6

$latex \displaystyle \int{\frac{3}{\sqrt{3u-5}},du}$

Problem 7

$latex \displaystyle \int{-4{{e}^{2p}},dp}$

Problem 8

$latex \displaystyle \int{5{{e}^{-0.3g}},dg}$

Do You Have More Integration By Parts Examples?

Here are some more examples of integration by parts courtesy of the face-to-face class.

Example 1

$latex displaystyle int{left( 2t-1 right)ln left( t right)dt}$















Example 2

$latex displaystyle int{{{x}^{4}}ln left( x right)dx}$






Example 3

$latex displaystyle int{ln left( 2x right)dx}$






Example 4

$latex displaystyle int{left( 5z-4 right)ln left( z right)dz}$





Example 5

$latex displaystyle int{{{x}^{3}}ln left( x right)dx}$
















Example 6

$latex displaystyle int{{{x}^{2}}ln left( x right)dx}$





 Example 7

$latex displaystyle int{ln left( 3z right)dz}$