Find the area of the shaded region below.

yes they are not so easy :)

\(Area=\\ -2\int_0^1(y^2-2)dy+\int_{-1}^1(e^y)dy\\ =-2*\left[\frac{y^3}{3}-2y\right]_0^1+\left[e^y\right] _{-1}^1\\ =-2*\left[(\frac{1}{3}-2)-(0-0)\right]+\left[e^1-e^{-1}\right] \\ =-2*\left[(-\frac{5}{3})\right]+\left[e-\frac{1}{e}\right] \\ =\frac{10}{3}+e-\frac{1}{e} \\ =\frac{10e+3e^2-3}{3e}\;\;units^2 \)

Ask if you have questions about what I have done :)

The cross-sectional area, A, is given by :

1

∫ [right function] - [left function]   dy 

-1 

So we have

1

∫     e^y  -  [ y^2  - 2] dy   =

-1

       1                    1                1

e^y]          -  y^3/3 ]       +   2y ]       =

     -1                    -1               -1

e - 1/e  -  [1/3 + 1/3]      +   4   =

10/3 + e - 1/e   ≈    5.6837  units^2

Find the area of the shaded region below.

\(\begin{array}{rcll} f(y) &=& e^y \\ g(y) &=& y^2-2 \\\\ A &=& \int \limits_{y=-1}^{y=1} { [f(y)-g(y)]\ dy}\\\\ A &=& \int \limits_{y=-1}^{y=1} { [ e^y-(y^2-2)]\ dy}\\\\ A &=& \int \limits_{y=-1}^{y=1} { ( e^y-y^2+2 )\ dy}\\\\ A &=& [~ e^y-\frac{y^3}{3}+2y ~]_{y=-1}^{y=1} \\ A &=& e^1-\frac{1^3}{3}+2\cdot 1 -(e^{-1}-\frac{(-1)^3}{3}+2\cdot(-1) ) \\ A &=& e -\frac13 +2 - ( \frac{1}{e}+\frac13 - 2) \\ A &=& e -\frac13 +2 - \frac{1}{e} - \frac13 + 2 \\ A &=& e - \frac{1}{e} -\frac23 + 4 \\ A &=& e - \frac{1}{e} + \frac{10}{3} \\ A &=& 2.71828182846 - 0.36787944117 + 3.33333333333 \\ \mathbf{A} & \mathbf{=} & \mathbf{ 5.68373572062 } \end{array}\)