Definite Intergrals problem

**Zanderist** · September 15th 2010, 02:21 PM

I'm only posting here because I am not seeing any way to do this in my Text book and my two math help books.

Let $\displaystyle \int_{10}^{19} f(x) dx =7, \ \int_{10}^{13} f(x) dx=9, \ \int_{16}^{19} f(x)dx =10$ .

Find $\displaystyle \int_{13}^{16} f(x)dx=$

and $\displaystyle \int_{16}^{13} (7 f(x)- 9)dx=$

I'm stuck on the first one.

I'm trying to use the Additive Interval Property as Shown Below.

$\mathbf{\int_{a}^{b}f(x) dx= \int_{c}^{a}f(x) dx + \int_{c}^{b}f(x) dx}$

Where a= 13, b=16.

So I try:

$\int_{10}^{13} f(x) dx + \int_{16}^{19} f(x)dx$

But keep ending up with the wrong answer, I've resorted by trying different combinations but with no success (even substitution).

**Danny** · September 15th 2010, 02:27 PM

Try $\displaystyle \int _{10}^{19} f(x)\,dx - \int_{16}^{19}f(x)\,dx - \int_{10}^{13}f(x)\,dx$ .

**Zanderist** · September 15th 2010, 02:39 PM

Well that worked...but why?

**pickslides** · September 15th 2010, 02:44 PM

Think about the required interval that is [13,16]

Danny's answer took a bigger interval [10,19] and cut off a bit at the bottom of this interval [10,13] and a bit at the top [16,19] leaving you with [13,16].

Do you follow?

**pickslides** · September 15th 2010, 02:50 PM

$\displaystyle \int_{16}^{13} (7 f(x)- 9)~dx$

$\displaystyle \int_{16}^{13} 7 f(x)~dx-\int_{16}^{13} 9~dx$

$\displaystyle 7\int_{16}^{13} f(x)~dx-\int_{16}^{13} 9~dx$

You can use the result to the first question to help finish this.

**Danny** · September 15th 2010, 02:55 PM

You have

$\displaystyle \int_{10}^{19} f(x)\,dx,\;\;\;\int_{10}^{13} f(x)\,dx,\;\;\;\int_{16}^{19} f(x)\,dx$

You want

$\displaystyle \int_{13}^{16} f(x)\,dx.$

$\displaystyle \int_{10}^{19} f(x)\,dx = \int_{10}^{13} f(x)\,dx + \int_{13}^{16} f(x)\,dx + \int_{16}^{19} f(x)\,dx$

Now isolate what you want (you have the other three integral values).

**Zanderist** · September 15th 2010, 03:05 PM

Originally Posted by pickslides

$\displaystyle 7\int_{16}^{13} f(x)~dx-\int_{16}^{13} 9~dx$

You can use the result to the first question to help finish this.

Okay well I have done that.

At first I tried.

7(-12)-9=-93

That was wrong.

So I then tried

7(-12)-9(-12)=24

And that was wrong.

I'll keep looking at this rule:

$\int_{a}^{b}[f(x)\pm g(x)]dx= \int_{a}^{b}f(x)dx \pm \int_{a}^{b} g(x)dx$

**pickslides** · September 15th 2010, 03:22 PM

Also $\displaystyle \int_{a}^{b}f(x)~dx = -\int_{b}^{a}f(x)~dx$

**Zanderist** · September 15th 2010, 03:31 PM

Just a quick question what's dx supposed to be?

Even with the rule you just said above.
I get the same answers as I did before only the sign is changed.

**Zanderist** · September 15th 2010, 07:14 PM

It's how to do the second part I can't figure out at the moment now.

I was looking at the wrong formula too.

I need to be looking at :

$\int_{a}^{b} kf(x)dx=k \int_{a}^{b} f(x)dx$

I'm starting to think that this has something to do with chain rule or compound functions (something along those lines).

**TheCoffeeMachine** · September 15th 2010, 11:40 PM

Originally Posted by Zanderist

Just a quick question what's dx supposed to be?

That the integration is with respect to $x$ .

I'm starting to think that this has something to do with chain rule or compound functions (something along those lines).

It's to do with the linearity of integration: if $k$ is any constant, then $\displaystyle \int_{a}^{b} kf(x)dx=k \int_{a}^{b} f(x)dx.$

It's called the constant factor rule of integration.

**Zanderist** · September 16th 2010, 04:25 PM

I've just solved it!

$\displaystyle \int_{16}^{13} (7 f(x)- 9)dx=$

$<br /> -7 \int_{16}^{13} f(-12) + 9\int_{16}^{13}dx$

The dx becomes b-a, or 16-13 = 3

so you end up with

-7(-12)+9(3)=111

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