Calculate the area between the two curves below?

Question

(Picture)

Answer 1

Probably easiest to Use  sideways  TE of Area .... that is integrate horizontally

 ∫ [ ( x value... on the right ) - ( x value... on the left ) ] dy  

   limits of integ. go from  lowest  to highest  y values in common...

      so   rt side    x + y = 2     thus     x  rt =  2 - y  ... similar for left side

     ∫  [ ( 2 - y )  - ( √ y )  ]  dy      limits  go  from  y = 0  to  y = 1

        [ 2y - (y^2) / 2  -  (2/3) y^(3/2) ]    from  0  to  1

   

         You  get    2 - (1/2) - (2/3)   =  5 / 6

or split  into two integrals  using vertical  Typical Elements  [ TE ]  of  area .....  vertical rectangles... 

Answer 2

I would break this up into two sections.

First, you want the area under the curve of y = x² from x = 0 to 1, then you want the area under the curve of x + y = 2 (or y = -x + 2) between x = 1 and 2.

The second one is easy as it's a triangle with a known base (1) and height (1), so the area of that is:

A = bh/2

A = 1(1)/2

A = 1/2 unit²

To get the area under the curve of the first section we need to integrate (anti-derivative) of the function.

To do that we can add one to the exponent and then divide the original coefficient by the new exponent to get the new coefficient.  There is normally a constant term here but since we are subtracting one integral from another the constant terms cancel out so we can ignore them.

y = x²

∫y = (1/3)x³

Now we solve this for x = 1 and subtract it from the value when x = 0:

(1/3)(1)³ - (1/3)(0)³

(1/3)(1) - (1/3)(0)

1/3 - 0

1/3 unit²

If we do the same thing to the triangle we will get the same area that we got using the area of a triangle equation, in case you wanted to give that a try for yourself, but I won't do that, here.

Finally, add the partial areas to get the total area:

1/2 + 1/3

3/6 + 2/6

5/6 unit²