Is My Integral Setup Correct for Calculating Area Between Curves?

In summary, the conversation revolves around setting up integrals and finding areas bounded by various functions. The first question involves finding the area between two given functions using integration, and the person is unsure if their answer is correct. The second question is about finding the area bounded by a line, a constant, and the x and y axes, and the third question is about finding the derivative of a natural logarithm function. The conversation includes questions and clarifications about how to set up the integrals and interpret the given information.
  • #1
gokugreene
47
0
Hey guys, I am curious if I am setting this up right. Could you take a look and make sure I am on the right path?

I have three questions.

1) I am trying to find the area bound by [tex]y=x^2[/tex] and [tex]y=4x+5[/tex]
Upper function [tex]y=4x+5[/tex]
Lower function [tex]y=x^2[/tex]
For my integral I have [tex]\int^{5}_{-1}4x-5-x^2[/tex]

I then get [tex]\left[\frac{-x^{3}}{3}+2x^{2}+5x\right]^{5}_{-1}[/tex]

and my answer is 110/3 units squared

Is my answer correct? Am I doing it right? I just figured out you can do it this way. My teacher has been teaching us to use squares, triangles, and other geometric methods (not talking about riemann sums) rather than just pure integration...

Ok, second question..
Find the area bounded by y=x-2, y=4, the x-axis, and the y-axis.
Would I set this problem up like [tex]\int^{6}_{2}x-2-\int^{2}_{0}x-2[/tex] this?

Last question..
I need to find the derivative of [tex]ln(x-1)^{2}[/tex]
I am confused as to whether they are saying to square x-1 or the natural log.

Thanks guys
 
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  • #2
On your first question you set up the problem correctly, but the integral of -5 is -5x, you put +5x, maybe that's a typo. I didn't check the exact answer, but you definitely have the right idea.

For the second question, remember that the area is the integral of the top minus the integral of the bottom. So set it up like this:

[tex]\int_{0}^{6} 4 - (x-2)dx[/tex]

They only want the first quadrant which is why the lower bound for the integral is zero.

As to the third question, that's rather ambiguous, I could interpret that either way.
 
  • #3
On the second question are they wanting me to find the area under x-2 or the area above it? It seems it could be looked at either way.
 
  • #4
gokugreene said:
On the second question are they wanting me to find the area under x-2 or the area above it? It seems it could be looked at either way.

From the graph and your bounds (x and y axes) it looks like you are trying to find the area in the first quadrant, above y=x-2 and below y=4. The area below x-2 is unbounded on the x axis, so your best bet is to go for the one that is contained the most.
 
  • #5
Alright, I will go with that. Thanks!
 
  • #6
ln(x-1)^2 take derivative, you must use chain rule
 

Related to Is My Integral Setup Correct for Calculating Area Between Curves?

What is the concept of "area between curves"?

The concept of "area between curves" refers to the calculation of the area enclosed by two or more curves on a graph. This can be done by finding the points of intersection between the curves and integrating the function that represents the difference between them.

How do I find the area between two curves?

To find the area between two curves, you must first determine the points of intersection between the curves. Then, set up an integral that represents the difference between the two functions. Finally, evaluate the integral to find the area enclosed by the curves.

What is the process for solving a math question involving the area between curves?

The process for solving a math question involving the area between curves is as follows: 1) Identify the curves and determine the points of intersection. 2) Set up an integral that represents the difference between the curves. 3) Evaluate the integral to find the area enclosed by the curves. 4) Check your answer by graphing the curves and the enclosed area.

Can the area between curves be negative?

Yes, the area between curves can be negative. This can occur when the lower curve is above the upper curve in certain regions, resulting in a negative value for the enclosed area. It is important to carefully consider the orientation of the curves when setting up and evaluating the integral.

What are some real-life applications of finding the area between curves?

One real-life application of finding the area between curves is in economics, where it can be used to calculate the profit or loss between two competing businesses. It is also used in physics to calculate the work done by a force and in engineering to determine the volume of a three-dimensional object. Additionally, the concept of area between curves is used in statistics to calculate the area under a probability curve.

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