The region R is the region in the first quadrant bounded by the curves y=x^ 2 -4x+ 4, x = 0 , and x = 2, as seen in the image attached below:

Question

ibb.co/WgrRRRL

Find a value h such that the vertical line x = h divides the region R into two regions of equal area. You get h = ?
I think the answer is 0.39987, but I could be wrong.

Help is greatly appreciated!

Answer 1

∫ ( x² - 4 x + 4 ) dx = x³ / 3 - 4 x² / 2 + 4 x + C

∫ ( x² - 4 x + 4 ) dx = x³ / 3 - 2 x² + 4 x + C

The area under the graph of a continuous function f(x) between the vertical lines x = a and x = b
can be computed by the definite integral:

b
∫ f(x) dx = F(b) - F(a)
a

In this case:

F(a) = F(0) = 0³ / 3 - 2 ∙ 0² + 4 ∙ 0 = 0

F(b) = F(2) = 2³ / 3 - 2 ∙ 2² + 4 ∙ 2 = 8 / 3 - 2 ∙ 4 + 4 ∙ 2 = 8 / 3 - 8 + 8 = 8 / 3

2
∫ ( x² - 4 x + 4 ) dx = F(2) - F(0) = 8 / 3 - 0 = 8 / 3
0

Now you must find the definite integral such as to be:

h
∫ ( x² - 4 x + 4 ) dx = F(h) - F(0) = 1 / 2 ( 8 / 3 ) = 8 / 6 = 2 ∙ 4 / 2 ∙ 3 = 4 / 3
0

F(h) = h³ / 3 - 2 h² + 4 h

F(0) = 0

F(h) - F(0) = F(h)

This means that you need to calculate:

F(h) = 4 / 3

h³ / 3 - 2 h² + 4 h = 4 / 3

Multiply both sides by 3

h³ - 6 h² + 12 h = 4

Now you must solve:

h³ - 6 h² + 12 h - 4 = 0

This equation can only be solved by numerical methods.

Using Newton-Raphson method:

x ≈ 0.41259

Answer 2

Thank you.