#### To determine

**To sketch:**

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

#### Answer

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

The graph with the typical approximating rectangle is

ii. The area of the region bounded by the given curve is A=43

#### Explanation

**1) Concept:**

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

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

Formula-

The area of the typical rectangle is

yT-yB∆x, where yT = top boundary curve and yB= bottom boundary curve.

The total area is

A= limn→∞∑i=1nyT-yB∆x = ∫ab(yT-yB) dx

**2) Given:**

y=x+3 and y=x+32

3) **Calculation:**

The given curves are y=x+3 and y=x+32

First, find the intersection of these two curves by solving the simultaneous equation.

x+3 =x+32

x+3-x+32=0

That is,

x+31-x+32=0

x+3=0 or 1=x+32

x=-3 or x=1

Therefore, the points of intersection of two curves is at x=-3, x=1

The graph of the region enclosed by the given curve is

Find the area by integrating with respect to x(since we found the top curve and the bottom curve from the graph and the restriction on x is given, so it is easy to find the area by integrating with the respective x)

From the graph, the top and the bottom curves are

yT=x+3 and yB=x+32

The graph with the typical rectangle is

The area of a typical rectangle is

yT-yB∆x=x+3-x+32∆x

And the region lies between x=-3 and x=1

So, the total area is

A= ∫ab(yT-yB) dx

= ∫-31x+3-x+32dx

To solve this integration, substitute

x+3=u

Thus, dx=du

x=-3⇒u=0 and x=1 ⇒u=4

Therefore, the integral becomes,

= ∫04u12-u2du

Integrating,

2u323-u2404

By using the upper and the lower limits,

=24323-1442-0

Simplifying,

=163-164

=163-4

A=43

Therefore, the area of the region bounded by the given curve is A=43

**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 curves is A=43