How do I solve this integral: ⌠(from -2 to 3) of |X+1|dx ?

[krete77]

How do I solve this integral: ⌠(from -2 to 3) of |X+1|dx ?

We did an example in class similar to this, but with the absolute value added on, it sort of confuses me. I'm looking for a STEP-BY-STEP break down so I can compare it with my notes from class.

I am taking Calc 1 and this is the end of the class (1 more left) and so he basically just threw this out at us, without to much explanation..Up until now, I've done ok, but really, I don't know where else to go. My academic resource center stinks and I can't find anything in my book (Calculus, briggs/cochran) to explain it.

If the problem confuses you, I can explain it here: it is the integral symbol, with -2 on the bottom of it, and 3 on the top, of the absolute value of X+1 dx. Thanks

Thanks in advance guys, please, a step by step process is what I am looking for

 

[micromass]

Hi krete77!

First, we will need to eliminate the absolute value. For this, you have to figure out when X+1\geq 0. Can you find out for which X that holds?

 

[krete77]

Yes, X=-1

 

[micromass]

For X=-1, we have that the equality holds, that is we have that X+1=0.
We want to know when X+1\geq 0.

 

[krete77]

Ok...so next?

 

[micromass]

Well, what is the answer? When is X+1 positive?

 

[krete77]

When X is positive, I have no idea..i'm really confused by this stuff, which is why I'm just asking for a complete STEP BY STEP explanation..(in the original post). Would you mind? I have an example here of integral of 2x-4 , copied the notes from class, i want to compare the 2..

 

[micromass]

So basically you want me to completely solve YOUR problem? Sorry, we don't do such a things here.

We are quite willing to help you in finding the solution, but you'll going to have to do the work.
The first thing we need to know is when X+1\geq 0, this is basic algebra, you shouldn't have much trouble with it.

We do we want to know that? Well, to eliminate the absolute value. If X+1\geq 0 then |X+1|=X+1 and if X+1\leq 0, then |X+1|=-X-1.

 

[krete77]

Ok, so then I integrate -(x+1) with the limits -2 to -1 + (x+1)dx with limits -1 to 3 . correct? from here, i need help integrating

 

[micromass]

Yes, that is correct.

It seems like you want to calculate the integral

\int{(x+1)dx}

how do you start this?

 

[krete77]

using the integral recipe; x^mdx=x^m+1/m+1; so I am in the process of doing this for both. which would give me: -x^2/2 evaluated from -2 to -1; + 1x evaluated from -2 to -1 + x^2/2 evaluated from -1 to 3 + 1x evaluated from -1 to 3.

Right? this is where i get lost

 

[micromass]

Indeed, so we have that

\int_{-2}^{-1}{(-x-1)dx}+\int_{-1}^3{(x+1)dx}=[-\frac{x^2}{2}-x]_{-2}^{-1}+[\frac{x^2}{2}+x]_{-1}^3

So all, we need to do know is to evaluate them. For example,

[-\frac{x^2}{2}-x]_{-2}^{-1}=(-\frac{(-1)^2}{2}-(-1))-(-\frac{(-2)^2}{2}-(-2))

So, now you just need to calculate this.

 

[krete77]

Sweet, so I actually told you the correct answer, in which I was confusing myself upon. Haha, thanks for the confirmation though, it really helped. So, now, after all calculations, the final answer is: 11 1/2

could you confirm this please, this is an even number in my book and I cannot double check my work. thanks for all your help and time. i can now rest from this semester peacefully ;)

 

[micromass]

No the correct answer should be 8.5
Recheck your calculations or post them here to see what went wrong.

 

[krete77]

i ran through my calculations again and got 11.5 still..here is what i have:

-((-1)^2/2)-(-1)-(-(-2)^2/2)-(-2)+((-1)^2/2)+(-1)+((3)^2)/2)+(3)=11.5

 

[micromass]

The calculation of the second part should be

-((-1)^2/2)-(-1)+((3)^2)/2)+(3)

So you needed to write - two times instead of a plus.

 

[SammyS]

i ran through my calculations again and got 11.5 still..here is what i have:

-((-1)^2/2)-(-1)-(-(-2)^2/2)-(-2)+((-1)^2/2)+(-1)+((3)^2)/2)+(3)=11.5

Use more grouping symbols to make it clearer:

[-((-1)^2/2)-(-1)]-[(-(-2)^2/2)-(-2)]+[((3)^2)/2)+(3)]-[((-1)^2/2)+(-1)]= ?

 

[krete77]

ah simple mistake, thanks again.