How to solve this integral of an absolute function?

In summary, the conversation discusses a problem involving finding the integral of a function involving absolute values. The function is even, so the integral from -1 to 1 is equal to 2 times the integral from 0 to 1. The conversation also discusses a graph and the incorrect use of the term k^2 in the solution.
  • #1
Helly123
581
20

Homework Statement


692wkm.jpg


Homework Equations

The Attempt at a Solution


I think the answer for number 1 , graph somewhat like this
w1pk7b.png


I get trouble for 2, 3, etc
I (k) = ##\int_{-1}^{1} f(x) dx ##
f(x) = ## \mid x^2 - k^2 \mid##

2) k < 1
for negative side
##\int_{-1}^{-k} (x^2 - k^2) dx + \int_{-k}^{0} (k^2 - x^2) dx ##
## \frac{x^3}{3} - k^2 {\mid} _{-1}^{-k} + (k^2 - \frac{x^3}{3} {\mid} )_{-k}^{0} ##
## \frac{-k^3}{3} - k^2 - (\frac{-1}{3} - k^2) + k^2 - (k^2 - \frac{-k^3}{3} ) ##
## \frac{1}{3} - \frac{2k^3}{3} ##

for positive side
##\int_{0}^{k} (k^2 - x^2) dx + \int_{k}^{1} ( x^2 - k^2 ) dx ##
## k^2 - \frac{k^3}{3} - k^2 + \frac{1}{3} - k^2 - (\frac{k^3}{3} - k^2) ##
## \frac{1}{3} - \frac{2k^2}{3} ##

total = ## - \frac{4k^3}{3} + \frac{2}{3} ##
but it is wrong.. why?

3) k > 1
for negative side
##\int_{-1}^{0} (k^2 - x^2) dx + \int_{0}^{1} (k^2 - x^2) dx ##
but I get zero.
 

Attachments

  • 692wkm.jpg
    692wkm.jpg
    10.2 KB · Views: 784
  • w1pk7b.png
    w1pk7b.png
    1.1 KB · Views: 784
Physics news on Phys.org
  • #2
$$
\int_a^b (x^2-k^2) dx \neq \left[\frac{x^3}{3} - k^2 \right]_a^b
$$
The ##k^2## term is incorrect.
 
  • Like
Likes Helly123
  • #3
Helly123 said:

Homework Statement


View attachment 221277

Homework Equations

The Attempt at a Solution


I think the answer for number 1 , graph somewhat like this
View attachment 221278

I get trouble for 2, 3, etc
I (k) = ##\int_{-1}^{1} f(x) dx ##
f(x) = ## \mid x^2 - k^2 \mid##

2) k < 1
for negative side
##\int_{-1}^{-k} (x^2 - k^2) dx + \int_{-k}^{0} (k^2 - x^2) dx ##
## \frac{x^3}{3} - k^2 {\mid} _{-1}^{-k} + (k^2 - \frac{x^3}{3} {\mid} )_{-k}^{0} ##
## \frac{-k^3}{3} - k^2 - (\frac{-1}{3} - k^2) + k^2 - (k^2 - \frac{-k^3}{3} ) ##
## \frac{1}{3} - \frac{2k^3}{3} ##

for positive side
##\int_{0}^{k} (k^2 - x^2) dx + \int_{k}^{1} ( x^2 - k^2 ) dx ##
## k^2 - \frac{k^3}{3} - k^2 + \frac{1}{3} - k^2 - (\frac{k^3}{3} - k^2) ##
## \frac{1}{3} - \frac{2k^2}{3} ##

total = ## - \frac{4k^3}{3} + \frac{2}{3} ##
but it is wrong.. why?

3) k > 1
for negative side
##\int_{-1}^{0} (k^2 - x^2) dx + \int_{0}^{1} (k^2 - x^2) dx ##
but I get zero.

You can simplify the analysis by noting that for any ##k## the function ##f(x) = |x^2 - k^2|## even (that is, ##f(-x) = f(x)##), so ##\int_{-1}^1 f(x) \, dx = 2 \int_0^1 f(x) \, dx##. You can see this from your graph, because you are trying to find the area under the graph from x = -1 to x = +1.

BTW: you are to be commended for now typing out your work---problem statements and solutions---instead of just posting images as you used to do. Why not go all the way and just dispense with the photos of the problem altogether? In this case they are so fuzzy as to be unreadable, and serve no purpose now anyway!
 
  • Like
Likes Helly123

1. What is an absolute function?

An absolute function, also known as an absolute value function, is a mathematical function that returns the positive value of a given number, regardless of its original sign. For example, the absolute value of -5 is 5.

2. How do I solve an integral of an absolute function?

To solve an integral of an absolute function, you can use the following steps:
1. Identify the limits of integration.
2. Rewrite the absolute function as a piecewise function.
3. Evaluate the integral for each piece of the function.
4. Add the results together to get the final solution.

3. Can I use the power rule to solve an integral of an absolute function?

Yes, you can use the power rule to solve an integral of an absolute function. However, you must first rewrite the absolute function as a piecewise function before applying the power rule.

4. Are there any special cases when solving an integral of an absolute function?

Yes, there are two special cases when solving an integral of an absolute function:
1. When the limits of integration are both positive or both negative, the absolute function can be rewritten as a regular function and the integral can be solved using the traditional methods.
2. When the limits of integration cross the origin (i.e. one limit is positive and the other is negative), the absolute function must be split into two pieces and the integral must be evaluated separately for each piece.

5. Can I use substitution to solve an integral of an absolute function?

Yes, substitution can be used to solve an integral of an absolute function. However, the substitution must be done carefully to ensure that the final solution accounts for all possible values of the absolute function. It may be easier to use the piecewise function approach in this case.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
429
  • Calculus and Beyond Homework Help
Replies
5
Views
178
  • Calculus and Beyond Homework Help
Replies
4
Views
589
  • Calculus and Beyond Homework Help
Replies
9
Views
675
  • Calculus and Beyond Homework Help
Replies
2
Views
454
  • Calculus and Beyond Homework Help
Replies
15
Views
735
  • Calculus and Beyond Homework Help
Replies
2
Views
453
  • Calculus and Beyond Homework Help
Replies
6
Views
776
  • Calculus and Beyond Homework Help
Replies
1
Views
364
  • Calculus and Beyond Homework Help
Replies
11
Views
3K
Back
Top