#### To determine

**To sketch:**

The region enclosed by the given curve and find the area of the region.

#### Answer

i. The graph of region enclosed by given curves is

The graph with the typical approximating rectangle is

ii. The area of region bounded by given curve is A=83

#### Explanation

**1) Concept:**

i. The intersection of the two curves is obtained by solving the simultaneous equation of the curves.

ii. Identify the top and the bottom boundaries of region.

Formula-

The area of the typical rectangle is

xR-xL∆y, where, RT= is the right boundary curve and xL is the left boundary curve.

The total area is

A= limn→∞∑i=1nxR-xL∆y = ∫abxR-xLdy

**2) Given:**

x=y2-1 and x=1-y2

3) **Calculation:**

The given curves are x=y2-1 and x=1-y2

First find the intersection of these two curve by solving the simultaneous equation:

y2-1=1-y2

y2-1-1+y2=0

that is,

2y2-2=0

2y2-1=0

y2-1=0

Simplifying,

y+1y-1=0

y=-1 or y=1

Therefore, the points of intersection of two curves are 0, -1, and (0, 1,)

The graph of the region enclosed by the given curves is

Find the area by integrating with respect to y(since, we found the left curve and the right curve from the graph and the restriction on y is calculated above, so it is easy to find area by integrating with the respective y).

From the graph, the leftmost and the rightmost curves are

xR=1-y2 and xL= y2-1

The graph with the typical rectangle is

The area of the typical rectangle is

xR-xL∆y=1-y2 -(y2-1) ∆y=(-2y2+2)∆y

And the region lies between y=-1and y=1

So, the total area is

A= ∫abxR-xLdy

= ∫-11(-2y2+2))dy

Integrating,

= -2y33+2y-11

By using the upper and the lower limits,

=-23+2-23-2

Simplifying,

=43+43

Simplifying,

A=83

Therefore, the area of the region bounded by the given curves is A=83

**Conclusion:**

i. The graph of the region enclosed by the given curves is

The graph with the typical rectangle is

ii. The area of the region bounded by the given curve is A=83