Expert: Paul Klarreich - 1/14/2010

Question
Hello paul, So my question is if there is a problems involving absolute value in these two examples how to i do them to get the answer. Could you run me through each step to find the answer because I have not seen an example of this yet.

1. [0,5]-could not do the integral sign. (abs(7x-6))dx=

On the next problem how do I find the answer given these other two answers, the problem looks like this:

2. Let *integral sign [10,19] f(x)dx=1, *integral sign [10,13] f(x)dx=8, *integral sign [16,19] f(x)dx=6.
They want me to find,
a. integral sign [13,16]f(x)dx=
b. integral sign [16,13]f(x)dx=

How do I find those with the given information? please run me thru the steps with the answer so when i run across future problems like this i'll understand it.

Thanks so much.

Answer
Questioner: Brooke
Country: United States
Category: Calculus
Private: No
Subject: integrals.
Question: Hello paul, So my question is if there is a problems involving absolute value in these two examples how to i do them to get the answer. Could you run me through each step to find the answer because I have not seen an example of this yet.

1. [0,5]-could not do the integral sign. (abs(7x-6))dx=

On the next problem how do I find the answer given these other two answers, the problem looks like this:

2. Let *integral sign [10,19] f(x)dx=1, *integral sign [10,13] f(x)dx=8, *integral sign [16,19] f(x)dx=6.
They want me to find,
a. integral sign [13,16]f(x)dx=
b. integral sign [16,13]f(x)dx=

How do I find those with the given information? please run me thru the steps with the answer so when i run across future problems like this i'll understand it.

Thanks so much.
...............................................


Here is what absolute value means:

abs(something) =

the something,                    if the something is >= 0  {not negative}

minus (opposite o) the something, if the something is < 0   {yes negative}

So the key in doing  | expression | is to find where the expression is zero, then split things.

To handle (abs(7x-6)), do this:

7x - 6 = 0  -->  x = 6/7.

So   when  x >= 6/7, use  7x - 6
and  when  x < 6/7, use  -(7x - 6), which is -7x + 6, of course.

Now then:  (BTW, this is how you make an integral sign.)

{ 5
|  abs(7x - 6) dx
} 0

must be split into two pieces:

{ 6/7
|  -(7x - 6) dx
} 0

{ 5
|  (7x - 6) dx
} 6/7

Do those two, add the answers, and you are done.

...............................................
2. Let *integral sign [10,19] f(x)dx=1, *integral sign [10,13] f(x)dx=8, *integral sign [16,19] f(x)dx=6.
They want me to find,
a. integral sign [13,16]f(x)dx=
b. integral sign [16,13]f(x)dx=

WHAT A MESS! OK, I think you mean:

{ 19
| f(x) dx = 1
} 10


{ 13
| f(x) dx = 8
} 10

{ 19
| f(x) dx = 6
} 16

Now the idea is this:

Integral from a to c = Integral from a to b + Integral from b to c,
where b is supposedly between a and c, but actually it does not have to be.

So (I will condense the notation now)

INT(10-19) = sum of three pieces: INT(10-13) + INT(13-16) + INT(16-19)

Think you can handle the rest?

And, of course:

INT(a-b) = - INT(b-a)