1. ## integrals

Let .

Find ________

and _________

2. Originally Posted by tbenne3
Let .

Find ________

and _________

Remember that

$\int_a^b{f(x)\,dx} = \int_a^c{f(x)\,dx} + \int_c^b{f(x)\,dx}$ for $a \leq c \leq b$.

So $\int_{-4}^{0.5}{f(x)\,dx} = \int_{-4}^{-2.5}{f(x)\,dx} + \int_{-2.5}^{-1}{f(x)\,dx} + \int_{-1}^{0.5}{f(x)\,dx}$.

3. Hello, tbenne3!

Did you make a sketch?

Let: . $\int^{0.5}_{-4}f(x)\,dx \:=\:4 \qquad \int^{-2.5}_{-4}f(x)\,dx\;=\;4 \qquad \int^{0.5}_{-1}f(x)\,dx \;=\;1$
These are areas "under" the curve.

With a little thought, we see that the graph looks like this:
Code:
                                      |
.*.                        |
.*:::::*.                     |
*::: 4 :::*               ..* *|.
:::::::::::           -1.*:: 1 |:*.
- - * - - - - - * - - - - - * - - - + - * - - - - -
-4        -2.5 *: -1 ::*         | -0.5
* *            |
|

$\text{Find: }\;\int^{-1}_{-2.5}f(x)\,dx$

$\text{Find: }\;\int^{-2.5}_{-1}\bigg[4f(x) - 4\bigg]\,dx$

Can you work out these areas now?

4. Originally Posted by Prove It
Remember that

$\int_a^b{f(x)\,dx} = \int_a^c{f(x)\,dx} + \int_c^b{f(x)\,dx}$ for $a \leq c \leq b$.

So $\int_{-4}^{0.5}{f(x)\,dx} = \int_{-4}^{-2.5}{f(x)\,dx} + \int_{-2.5}^{-1}{f(x)\,dx} + \int_{-1}^{0.5}{f(x)\,dx}$.
so where do I go from there?

5. Originally Posted by Soroban
Hello, tbenne3!

Did you make a sketch?

These are areas "under" the curve.

With a little thought, we see that the graph looks like this:
Code:
                                      |
.*.                        |
.*:::::*.                     |
*::: 4 :::*               ..* *|.
:::::::::::           -1.*:: 1 |:*.
- - * - - - - - * - - - - - * - - - + - * - - - - -
-4        -2.5 *: -1 ::*         | -0.5
* *            |
|

Can you work out these areas now?

I don't understand what it's asking for

6. Originally Posted by tbenne3
so where do I go from there?
You should realise that you've been given 3 of those 4 integrals. So substitute them and solve for the 4th.

7. Originally Posted by Prove It
You should realise that you've been given 3 of those 4 integrals. So substitute them and solve for the 4th.
I got the first but must be doing something wrong for the second

8. Well that's not surprising considering that I haven't taught you how to do Part 2 yet...

$\int_{-1}^{-2.5}{4f(x) - 4\,dx} = \int_{-1}^{-2.5}{4f(x)\,dx} - \int_{-1}^{-2.5}{4\,dx}$

$= 4\int_{-1}^{-2.5}{f(x)\,dx} - \int_{-1}^{-2.5}{4\,dx}$

$= -4\int_{-2.5}^{-1}{f(x)\,dx} + \int_{-2.5}^{-1}{4\,dx}$.

9. Originally Posted by Prove It
Well that's not surprising considering that I haven't taught you how to do Part 2 yet...

$\int_{-1}^{-2.5}{4f(x) - 4\,dx} = \int_{-1}^{-2.5}{4f(x)\,dx} - \int_{-1}^{-2.5}{4\,dx}$

$= 4\int_{-1}^{-2.5}{f(x)\,dx} - \int_{-1}^{-2.5}{4\,dx}$

$= -4\int_{-2.5}^{-1}{f(x)\,dx} + \int_{-2.5}^{-1}{4\,dx}$.
can't figure out why i'm not getting it right.. what are you plugging in for f(x)?