#### To determine

**To evaluate:**

The integral and interpret it as the area of a region.

#### Answer

i) The value of the integral is

443-46-432

ii) The sketch of the region

#### Explanation

**1) Concept:**

Formula:

The area A of the region bounded by the curves y=f(x), y=g(x) and the lines x=a and x=b is

A= ∫abfx-gxdx

fx-gx=fx-gx when fx≥g(x)gx-fx when gx≥f(x)

**2) Given:**

∫04x+2-xdx

3) **Calculation:**

Comparing the given integral with A= ∫abfx-gxdx gives,

fx=x+2 and gx=x

To find the intersection point of the curves, equate f(x) and gx with each other simultaneously on the interval 0, 4.

Therefore, x+2=x

Square both sides.

x+22=x2

x+2=x2

x2-x-2=0

x=2 or x=-1

But, -1 is not within the given interval 0, 4.

Therefore, the curves intersect at x=2

Splitting the given interval at x=2 gives,

0, 2 and 2, 4

In the interval 0, 2, x-x+2 ≤0

I.e. x≤x+2

Therefore,∫02(x+2-x)dx=23x+23/2-12x220

=232+232-1222-230+232-1202

=103-423

In the interval 2, 4, x-x+2≥0

i.e. x≥x+2

Therefore,∫24(x-x+2)dx=12x2-23x+23/242

=1242-234+23/2-1222-232+23/2

=343-46

Now, combining both the areas, it becomes

∫04x+2-xdx=∫02(x+2-x)dx+∫24(x-x+2)dx

=443-46-432

Between 0, 2, the upper curve is y=x+2 and the lower curve is y=x

And over the interval [2, 4], the upper curve is y=x and the lower curve is y=x+2

The total area

=443-46-432

The sketch of the region

**Conclusion:**

i) The total area of the region using the integral is

443-46-432

ii) The sketch of the region