Hello, Apprentice123!
Did you sketch the graph?
Calculate the area by integration of the region limited by the curves:
. . . $\displaystyle y\:=e\:^{x},\quad y \:=\:x+1,\quad x\:=\:1$
Answer: .$\displaystyle e\frac{3}{2}$ Code:
* 

*  *
*  *
*  *
::*  *
:::::*
:::*  *
:*  *
   *   +         
* 
* 
$\displaystyle 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} $
. . $\displaystyle = \;\;\bigg[e^0  0  0\bigg]  \bigg[e^{\text{}(\text{}1)}  \frac{1}{2}(\text{}1)^2  (\text{}1)\bigg]$
. . $\displaystyle = \;\;1  \left(e  \frac{1}{2} + 1\right) \;\;=\;\;1 + e  \frac{1}{2} \;\;=\;\;e  \frac{3}{2}$
I doubt if anyone will help you with the other two problems
. . until you show some effort.