`y=x`

`y=4`

`x=0`

The region bounded by the three equations is:

To determine the area of the bounded region, draw a vertical strip. (See attached image.)

In the figure, the top of the vertical strip touches the graph of y=4. And its lower end touches the graph of y=x. Also, the bounded region starts from x=0 and ends at x = 4.

So, applying the formula

`A = int_a^b (y_(_(upper)) - y_(_(lower)))dx`

the integral needed to compute the area of the bounded region is:

`A = int_0^4 (4 - x)dx`

Evaluating it results to:

`A = (4x - x^2/2)` `|_0^4`

`A = (4*4 - 4^2/2) - (4*0 - 0^2/2)`

`A = 8`

Hence, the area of the region bounded by the three equations is 8 square units.

Then, draw a vertical line that divides the area of the enclosed region equally. Let the equation of this line be x=a. (See attached image.)

Since it divides the area equally, the area of the region at the left of the vertical line is equal to the area of the region at the right. And the area of each is 4 square units.

`A_(_(l eft))= A_(_(r ight)) = A/2 = 8/2=4`

Considering the region at the left of x=a, the integral needed to compute its area is:

`A_(_(l eft)) = int _0^a (4-x)dx`

Evaluating the integral, it results to:

`A_(_(l eft)) = (4x - x^2/2) |_0^a`

`A_(_(l eft)) = (4*a - a^2/2) - (4*0 - 0^2/2)`

`A_(_(l eft)) = 4a-a^2/2`

To determine the value of a, plug-in the area of the region at the left of the vertical line.

`4=4a - a^2/2`

Take note that to solve quadratic equation, one side should be zero.

`a^2/2 - 4a + 4 = 0`

To eliminate the fraction in the equation, multiply both sides by 2.

`2*(a^2/2 - 4a + 4) = 0*2`

`a^2-8a+ 8a = 0`

Applying quadratic formula, the values of a are:

`a = (-(-8)+-sqrt((-8)^2-4*1*8))/(2*1)`

`a=(8+-sqrt32)/2`

`a=(8+-4sqrt2)/2`

`a=4+-2sqrt2`

`a_1=4-2sqrt2=1.1716`

`a_2=4+2sqrt2 = 6.8284`

Between them, it is only a1=1.1716 that is within the interval of the bounded region.

**Therefore, the equation of the vertical line that divides the area of the bounded region equally is `x = 1.1716` .**

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