Evaluating Integrals: Additive Interval Property

  • Thread starter mathpat
  • Start date
  • Tags
    Integrals
In summary, the conversation is about solving an integral problem involving two given equations and finding the area between specific intervals. The correct solution is 11. The approach used was the additive interval property of integrals.
  • #1
mathpat
15
0

Homework Statement




Given
7 f(x) dx= 8
0

7 f(x) dx = −3
1

evaluate the following.

1 f(x) dx
0



Homework Equations


n/a


The Attempt at a Solution



I'm a little confused on how to approach this problem. Do i use the additive interval property of integrals?
 
Physics news on Phys.org
  • #2
mathpat said:

Homework Statement




Given
7 f(x) dx= 8
0

7 f(x) dx = −3
1

evaluate the following.

1 f(x) dx
0



Homework Equations


n/a


The Attempt at a Solution



I'm a little confused on how to approach this problem. Do i use the additive interval property of integrals?
Yes.
 
  • #3
I ended up with 5. Is that correct?
 
Last edited:
  • #4
mathpat said:
I ended up with 5. Is that correct?
So 5 + (-3) = 8?
 
  • #5
To expand on what Mark44 is saying, remember that these integrals represent areas under your curve, f(x). If you know how much area is under the curve between x = 0 and x = 7 and also know how much area is under the curve between x = 1 and x = 7, can you intuitively decide how to find the area between x = 0 and x = 1?
 
  • #6
Yea I understand but when I use the formula i keep getting 5. I don't see where I'm going wrong.
 
  • #7
[itex] \int_0^7 f(x) dx = \int_0^1 f(x) dx + \int_1^7 f(x) dx [/itex]. What happens when you plug in what you know?
 
  • #8
I calculated 11 using that formula.
 
  • #9
Good, that's right.
 
  • #10
Thanks
 

1. What is the Additive Interval Property for evaluating integrals?

The Additive Interval Property states that the integral of a function over a sum of two intervals is equal to the sum of the individual integrals over each interval.

2. How is the Additive Interval Property used to evaluate integrals?

The Additive Interval Property is used by breaking down the integral into smaller intervals and evaluating each one separately. Then, the results are added together to get the final integral value.

3. Can the Additive Interval Property be applied to any type of function?

Yes, the Additive Interval Property can be applied to any integrable function. However, it is most commonly used for piecewise functions or functions that have discontinuities at certain points.

4. What is the difference between the Additive Interval Property and the Distributive Property?

The Additive Interval Property is specific to integrals and involves breaking down the integral into smaller intervals. The Distributive Property, on the other hand, is a general mathematical property that applies to operations like multiplication and addition.

5. Are there any limitations to using the Additive Interval Property for evaluating integrals?

One limitation of the Additive Interval Property is that it only applies to integrals over intervals. It cannot be used for integrals over non-interval sets, such as circles or ellipses. Additionally, there may be cases where breaking down the integral into smaller intervals may not be possible or may result in a more complex integral to evaluate.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
420
  • Calculus and Beyond Homework Help
Replies
9
Views
543
  • Calculus and Beyond Homework Help
Replies
9
Views
756
  • Calculus and Beyond Homework Help
Replies
10
Views
438
  • Calculus and Beyond Homework Help
Replies
4
Views
650
  • Calculus and Beyond Homework Help
Replies
2
Views
383
  • Calculus and Beyond Homework Help
Replies
2
Views
841
  • Calculus and Beyond Homework Help
Replies
8
Views
660
  • Calculus and Beyond Homework Help
Replies
1
Views
762
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top