# Thread: i dunno how to do these integral problems

1. ## i dunno how to do these integral problems

there are 2 that i dunno how to do.

Re:

3. ## for the 1st problem

for the 1st problem

4. Hello, falloutdude!

Did you make a sketch for the second problem?

Let: .$\displaystyle f(x) \;=\;\begin{Bmatrix} 0 & \text{if }x< 0 \\ x & \text{if }0 \leq x\leq 3 \\ 6-x & \text{if }3< x\leq 6 \\ 0 & \text{if }x > 6\end{Bmatrix}$ . and: .$\displaystyle g(x) \:=\:\int^x_0f(t)\,dx$

Find an expression for $\displaystyle g(x)$ when $\displaystyle 3 < x <6.$
The graph looks like this:
Code:
              |
|       (3,3)
|         *
|       *:|:*
|     *:::|:::*
|   *:::::|:::| *
| *:::::::|:::|   *
- - * * * * - - - - + - + - - * * * * - -
0         3   x     6
$\displaystyle g(x)$ is the area under the graph from $\displaystyle 0$ to $\displaystyle x.$

The left region is a right triangle with base 3 and height 3; its area is $\displaystyle \frac{9}{2}$

The right region is the area under $\displaystyle y \:=\:6 - x$ from $\displaystyle 3$ to $\displaystyle x.$
. . This is: .$\displaystyle \int^x_3(6 - x)\,dx \;=\;6x-\frac{1}{2}x^2\bigg]^x_3 \;=\;6x - \frac{1}{2}x^2 - \frac{27}{2}$

Therefore: .$\displaystyle g(x) \;=\;\frac{9}{2} + \left(6x -\frac{1}{2}x^3 - \frac{27}{2}\right)\quad\Rightarrow\quad\boxed{g(x )\;=\;6x - \frac{1}{2}x^2 - 9}$

5. ## the graph of f(x)

Originally Posted by Soroban
Hello, falloutdude!

Did you make a sketch for the second problem?

The graph looks like this:
Code:
              |
|       (3,3)
|         *
|       *:|:*
|     *:::|:::*
|   *:::::|:::| *
| *:::::::|:::|   *
- - * * * * - - - - + - + - - * * * * - -
0         3   x     6
$\displaystyle g(x)$ is the area under the graph from $\displaystyle 0$ to $\displaystyle x.$

The left region is a right triangle with base 3 and height 3; its area is $\displaystyle \frac{9}{2}$

The right region is the area under $\displaystyle y \:=\:6 - x$ from $\displaystyle 3$ to $\displaystyle x.$
. . This is: .$\displaystyle \int^x_3(6 - x)\,dx \;=\;6x-\frac{1}{2}x^2\bigg]^x_3 \;=\;6x - \frac{1}{2}x^2 - \frac{27}{2}$

Therefore: .$\displaystyle g(x) \;=\;\frac{9}{2} + \left(6x -\frac{1}{2}x^3 - \frac{27}{2}\right)\quad\Rightarrow\quad\boxed{g(x )\;=\;6x^2 - \frac{1}{2}x^2 - 9}$
the graph of f(x)

6. ## and the graph of g(x)

and the graph of g(x)