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Integral & Piecewise Function

  • Thread starter Jet1045
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  • #1
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Hey everyone, alright so i have two calculus questions that i need help with:

1.
the Integral (lower bound 'a' , upper bound 'b') of f(x) dx = a + 2b

Then

the Integral (lower bound 'a' , upper bound 'b') of (f(x) + 3) dx = ?

I have no idea what the second integral would equal, knowing that the first one equals a + 2b. Since in the second integral you are adding 3 to f(x), would it just equal a + 2b + 3?
Obviously that seems too simple, so if someone could help thatd be great!

--------

2. I was given the Piecewise Defined function of:
f(x) =
-1 when x < 0
0 when x = 0
1 when x > 0

One of the questions following , was :

What is the limit as x approaches 0 of the absolute value of f(x)?

Would the answer be 0, 1 or Does Not Exist?

If someone could help that'd be great, as I have extra time to finish the test tomorrow. haha

THANKS! :)
 

Answers and Replies

  • #2
CompuChip
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There are two ways to solve this, one is by algebra, the other by geometry.
Can you explain what [itex]\int_a^b f(x) \, dx[/itex] means (for example, in terms of the graph of f)?
And can you tell me how the graph of f(x) + 3 is related to that of f(x)?

For the second one, can you give a definition of the absolute value of f(x), in the same way as of f(x). I.e. |f(x)| = ... if x < 0, ....etc
 
  • #3
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I think my teacher is looking for the algebraic way ,
and honestly i have no other information than that :( that is exactly what he stated as the question. We were not given a graph to reference, only f(x) and f(x) + 3. I guess he is just using a generic function, and then the same function with an added 3 to the end .

----

For the second one, f(x) equals that entire piecewise function, so it could be the absolute value of any x value. I was able to draw out the graph, but am unable to tell if the limit is 1, 0 or DNE.
sorry that I can't really explain these things further, I am not really doing well in this class, (well for me anyway) :(
 
  • #4
SammyS
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2. I was given the Piecewise Defined function of:
f(x) =
-1 when x < 0
0 when x = 0
1 when x > 0

One of the questions following , was :

What is the limit as x approaches 0 of the absolute value of f(x)?

Would the answer be 0, 1 or Does Not Exist?

If someone could help that'd be great, as I have extra time to finish the test tomorrow. ha ha

THANKS! :)
What are: lim x→0+ |f(x)| and lim x→0 |f(x)| ?
 
  • #5
Char. Limit
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I think my teacher is looking for the algebraic way ,
and honestly i have no other information than that :( that is exactly what he stated as the question. We were not given a graph to reference, only f(x) and f(x) + 3. I guess he is just using a generic function, and then the same function with an added 3 to the end .
Just use the sum of integrals property:

[tex]\int_a^b (f(x)+g(x)) dx = \int_a^b f(x) dx + \int_a^b g(x) dx[/tex]

Let f(x) = f(x) and g(x) = 3.
 
  • #6
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0
Thank you soooo much Char. Limit , i honestly don't know how I didn't see that... haha

so the anti-derivative of 3 would be 3x, so then would it make sense for me to go: 3(b) - 3(a)
and then i would add the two integral results together right:

a + 2b + 3b - 3a
= -2a + 5b

let me know if this makes sense :)

------


SammyS, I am not sure what those limits would me, as I have never done absolute value limits with a Piecewise function.
I know that the
lim as x approaches 0- = -1 &
lim as x approaches 0+ = 1

But am unsure as how to apply an absolute value to that function.
I can easily graph it and see how it works, but have no clue how to apply absolute values at all :(
 
  • #7
Char. Limit
Gold Member
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13
Thank you soooo much Char. Limit , i honestly don't know how I didn't see that... haha

so the anti-derivative of 3 would be 3x, so then would it make sense for me to go: 3(b) - 3(a)
and then i would add the two integral results together right:

a + 2b + 3b - 3a
= -2a + 5b

let me know if this makes sense :)
Makes sense to me.
 
  • #8
49
0
Makes sense to me.
Thanks so much. hey quick question, if you can answer i dont have to make a new thread.
i am doing logrithmic differentiation. so i have a y = equation, and i have taken the 'ln' of both sides, after i take the ln of each term in the equation, one of the terms is:
lne^((x^2) + 2)

to get the derivative, do i go
(1/e^((x^2) + 2)) multiplied by the derivative of the inner function which is 2x

OR would i bring the ((x^2) + 2) in front of lne
making it:

((x^2) + 2)lne and then use the product rule to get:

2x

i am really confused with this! any help would be great.

I just dont know if am an supposed to bring down the exponent of 'e' in front of the lne and use product rule, or leave it as be and use the chain rule.
 

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