1. ## Another Calc Question...

Q:

2. Originally Posted by qbkr21
Q:
You know that
$\int_{-1}^{6.5} dx \, f(x) = 10$
$\int_{-1}^{1.5} dx \, f(x) = 5$
and
$\int_4^{6.5} dx \, f(x) = 7$

So
$\int_{1.5}^4 dx \, f(x) = \int_{-1}^{6.5} dx \, f(x) - \int_{-1}^{1.5} dx \, f(x) - \int_4^{6.5} dx \, f(x)$

$= 10 - 5 - 7 = -2$

Thus
$\int_4^{1.5} dx \, f(x) = - \int_{1.5}^4 dx \, f(x) = 2$

And we know that
$\int_4^{1.5} dx \, (-5) = -5(1.5 - 4) = 12.5$

So finally:
$\int_4^{1.5} dx \, (10f(x) - 5) = 10(2) + 12.5 = 32.5$

-Dan

3. ## Re:

Thanks,

-qbkr21

4. Originally Posted by qbkr21
Q:
i cant see if there are decimal places for some of those limits, i suppose the top limits are 1.5 and 6.5

Given: $\int_{-1}^{6.5} f(x) dx = 10$ , $\int_{-1}^{1.5} f(x) dx = 5$, and $\int_4^{6.5} f(x) dx = 7$

Now, $\int_{1.5}^{6.5} f(x) dx = \int_{-1}^{6.5} f(x) dx - \int_{-1}^{1.5} f(x) dx = 10 - 5 = 5$

Also, $\int_4^{1.5} f(x) dx = \int_4^{6.5} f(x) dx - \int_{1.5}^{6.5} f(x) dx = 7 - 5 = 2$

Now we are ready to solve the problem:

$\int_4^{1.5} (10f(x) - 5)dx = 10 \int_4^{1.5} f(x) dx - \int_4^{1.5} 5dx$

I think you can take it from here

EDIT: Beaten by Dan yet again. I did it just a little bit different though. Just you wait till i get fast with Latex Dan

5. ## Re:

Q:

6. Originally Posted by qbkr21
Q:
that is incorrect. no one said that.

the idea is we want to get rid of what we don't want and leave what we do. it is the same principle as with the addtion. it's like if we want the area of a disk with a hole in it, we would find the area of the disk and subtract the area of the whole from it

if you take the equations where you saw Dan and I had one integral minus another and add the minus integral to both sides, you will realize you have the same addtion relationship you described above. that's the underlying principle in what we did, we just never wrote it

for instance, when i had $\int_4^{1.5} f(x) dx = \int_4^{6.5} f(x) dx - \int_{1.5}^{6.5} f(x) dx$

i could have written $\int_4^{1.5} f(x) dx + \int_{1.5}^{6.5} f(x) dx = \int_4^{6.5} f(x) dx$

which is the relationship you said you know about

7. qbkr21:
I'd quote your last message but it doesn't appear in the quote!

For the record, for a < b < c:
$\int_a^c dx \, f(x) = \int_a^b dx \, f(x) + \int_b^c dx \, f(x)$

Thus
$\int_b^c dx \, f(x) = \int_a^c dx \, f(x) - \int_a^b dx \, f(x)$

It's just subtraction.

-Dan