# Integration by parts and negatives

1. Feb 9, 2012

### bobsmith76

1. The problem statement, all variables and given/known data

Here are two instances where the negative sign just changes for no reason. The one's all the way on the right. Why? I don't understand what is going on here. For the second one, it should + cos x

2. Feb 9, 2012

### Vonalak

The negative sign was carried when it was distributed (negative * negative = positive). There's no mistake. Glad to help.

3. Feb 9, 2012

### SammyS

Staff Emeritus
It's not so much that it changes for no reason. It's just that you don't see why it changes.

What is a + b - (c -d) ?

It's a + b - c + d , correct ?

Be more accepting of the results in your textbook and try to resolve these issues on your own. Once you do, you will have a much better grasp of what's going on than if you quickly turn to others for help immediately upon finding a result you don't understand.

Trying to understand Calculus is extra difficult when you don't have a firm grasp of Algebra.

4. Feb 9, 2012

### bobsmith76

Of course, I'm aware of the distributive law. I still don't see what's going on.

In the first one, they go from

- ∫ (sin x)(6 dx)

to

+ ∫ (6sin x)

It doesn't make sense.

In the second one they go from

- ∫ - cos x ex dx

to

- ∫ ex cos x dx

Sammy,

If I see + + = - , then how am I supposed to resolve that on my own?

5. Feb 9, 2012

### SammyS

Staff Emeritus
Where is + + = - ?

a - ( -b - (-c)) = a + b - c .

6. Feb 9, 2012

### bobsmith76

essentially what i see is a violatation of the distributive law

i see (-sinx) * (6) = 6sinx

and

- - cos x ex = - cos x ex

in other words, -a * a = a, and - -a * a = -a

7. Feb 9, 2012

### SammyS

Staff Emeritus
Let's do this in smaller steps. (This is a good idea when you don't see what's happening.)

$\displaystyle \dots \ \ -\left((6x)(\sin\,x)-\int(\sin\,x)(6\,dx) \right)$

$\displaystyle =\ \dots \ \ +(-1)\left(6x\sin\,x-\int(6\sin\,x\,dx) \right)$

$\displaystyle =\ \dots \ \ +\left(-6x\sin\,x+\int(6\sin\,x\,dx) \right)$

$\displaystyle =\ \dots \ \ -6x\sin\,x+\int(6\sin\,x)\,dx$

8. Feb 9, 2012

### bobsmith76

$\displaystyle \dots \ \ -\left((6x)(\sin\,x)-\int(\sin\,x)(6\,dx) \right)$

$\displaystyle =\ \dots \ \ +(-1)\left(6x\sin\,x-\int(6\sin\,x\,dx) \right)$

$\displaystyle =\ \dots \ \ +\left(-6x\sin\,x [i don't see why this changes to positive]+\int(6\sin\,x\,dx) \right)$

$\displaystyle =\ \dots \ \ -6x\sin\,x+\int(6\sin\,x)\,dx$

9. Feb 9, 2012

### SammyS

Staff Emeritus
You don't see why $\displaystyle (-1)\left(6x\sin\,x-\int(6\sin\,x\,dx) \right)=-6x\sin\,x+\int(6\sin\,x)\,dx\,? \ ? \ ?$

That's the distributive law.

10. Feb 9, 2012

### bobsmith76

Ok, so you're multiplying -1 by every term within the parentheses. Now I got it.

11. Feb 9, 2012

### Vonalak

As someone said earlier, without a strong foundation in algebra, calculus will be difficult, especially at this level. The calculus can sometimes get hard enough as is, without the tricky minutiae of algebra!

12. Feb 10, 2012

### bobsmith76

Doing Algebra is boring. I've tried it. It's too boring. Doing calculus you do algebra and calculus at the same time. the first half of the problem is the calc part, the second half is the algebra part, much more exciting.

13. Feb 10, 2012

### Curious3141

Well, boring or not, you're going to have to get to grips with algebra before you can get anywhere with calculus. This is basically the advice I gave you in an earlier thread, and I'm reiterating it.

14. Feb 10, 2012

### bobsmith76

the mistake above was a calculus mistake not an algebra mistake. i didn't understand where the negative 1 came from. as for not distributing the negative to every term within the parenthesis, that's the equivalent to a typo, not a misunderstanding of algebra.

15. Feb 10, 2012

### micromass

Staff Emeritus
The mistake above was a pure basic algebra mistake. Seriously, if you don't see right away why

$$-(b-c)=-b+c$$

then calculus is perhaps not for right now. Start by reviewing basic algebra and trig. Calculus books will not go over basic algebra for you and will assume that you know it. You seriously can't do calculus without knowing the distributive law!!!!