# Math Help - Area 1

1. ## Area 1

Calculate the area by integration of the region limited by the curve: And build the graph

$y=e^(-x)$, $u=x+1$ and $x=-1$

And build the graph

$e-\frac{3}{2}$

2. Hello, Apprentice123!

Did you sketch the graph?

Calculate the area by integration of the region limited by the curves:

. . . $y\:=e\:^{-x},\quad y \:=\:x+1,\quad x\:=\:-1$

Answer: . $e-\frac{3}{2}$
Code:
        *       |
|
*      |       *
*     |     *
|*    |   *
|::*  | *
|:::::*
|:::* |   *
|:*   |           *
- - - * - - + - - - - - - - - -
*       |
*         |

$A \;=\;\int^0_{\text{-}1}\bigg[e^{-x} - (x + 1)\bigg]\,dx \;=\;\int^0_{\text{-}1}\bigg[e^{-x} - x - 1\bigg]\,dx \;\;=\;\;-e^{-x} - \frac{1}{2}x^2 - x\,\bigg]^0_{\text{-}1}$

. . $= \;\;\bigg[-e^0 - 0 - 0\bigg] - \bigg[-e^{\text{-}(\text{-}1)} - \frac{1}{2}(\text{-}1)^2 - (\text{-}1)\bigg]$

. . $= \;\;-1 - \left(-e - \frac{1}{2} + 1\right) \;\;=\;\;-1 + e - \frac{1}{2} \;\;=\;\;e - \frac{3}{2}$

I doubt if anyone will help you with the other two problems
. . until you show some effort.

3. which the points of intersection?

4. You didn't sketch the graph, did you?

Did you look at my diagram?
Can't you FIND the intersections?