Write an integral

Question

Let the region R be the area enclosed by the function f(x) = ln(x) +2 and g(x) =4/3x. Write an integral in terms of x and also in terms of y that would represent the region R If necessary round limit values to nearest thousandth.

Answer 1

desmos.com/calculator/fvmjurl3cx

Answer 2

Given: Region R, enclosed by f(x) = ln(x) + 2 and g(x) = 4/3x

Q: Write an integral in terms of x and also in terms of y that would represent the region R

Before we dive in, an interval in terms of “x” and in terms of “y” will be 2 different integrals, one integrating vertical partitions along the x-axis, over values of [f(x) - g(x)].

When integrating over ‘y’, we partition along the y-axis, with rectangles lengths along the x-axis, between the 2 curves. For this we need to express x in terms of y via the inverse function for both f() and g().

y = f(x) = ln(x) + 2 ==> y - 2 = ln(x) ==> x = fᵧ(y) = e⁽ʸ ⁻ ²⁾

y = g(x) = (4/3)x ==> x = gᵧ(y) = (3/4)y

Region R

…is defined by points of intersection… when f(x) = g(x) ==> ln(x) + 2 = (4/3)x

Do we really need to find the points of intersection ???

Well, you cannot write Definite Integrals unless you know the range of integration, right ?

ln(x) + 2 = (4/3)x : smells a bit related to ln(x)/x = … We know that ln(u)/u has max value = 1/e at u=e, so it could be useful to investigate powers of ‘e’ for near intersecting values.

I’ll use the Newton-Raphson Method (aka Newton’s Method) Approximation Method to find the precise intersections points for our range of Integration…

We need to find the zeroes of F(x) = f(x) - g(x) = (ln(x) + 2) - (4/3)x, because this is when the 2 curves intersect. We cannot setup any Integrals for area of R unless we have a range of Integration.

Basic idea here: Study the 2 curves, estimate 2 points “near” each intersection… emphasis on “near”… so study the curves.

Since y = (4/3)x passes through the origin, and we have a “feel” for y = ln(x) + 2, we know there should be an intersection very close to the Origin, and another one "farther out”.

Then the Newton-Raphson Method performs “iteration” on a suitably chosen initial value to converge… very rapidly !

a. F(1) = 0 + 2 - 4/3 = 2/3 > 0 # not far out enough !

b. F(e¹) = 1 + 2 - (4/3)(e) = -0.62 # we overshot the intersection, between 1 and e… go back a bit !

c. F(e¹/²) = 1/2 + 2 - (4/3)(e¹/²) = 0.3 # overshot, go back a bit

==> x = e³/⁴ ≈ 2.117 is of interest… F(e³/⁴) = 3/4 + 2 - (4/3)e³/⁴ = -0.07 # Reasonable starting point

Now, for the intersection closer to the Origin...

a. F(1/100) = -4.61 + 2 - (4/3)(1/100) = -2.62 # ln(x) + 2 is way below our line (4/3)x

b. F(1/10) = -2.3 + 2 - (4/3)(1/10) = -0.43 # still below, go forward

c. F(1/5) = -1.61 + 2 - (4/3)(1/5) = 0.12 <==

d. F(1/3) = -1.1 + 2 - (4/3)(1/3) = -0.45

e. F(1/4) = -1.39 + 2 - (4/3)(1/4) = 0.27 <==

f. F(0.2) = -1.61 + 2 - (4/3)(.2) = 0.12

g. F(0.18) = -1.715 + 2 - (4/3)(0.18) = 0.045

h. F(0.15) = -1.898 + 2 - (4/3)(0.15) = -0.098

i. F(0.16) = -1.833 + 2 - (4/3)(0.16) = -0.046

j. F(0.17) = -1.772 + 2 - (4/3)(0.17) = 0.0013 <===

==> x = 0.17 is of interest, as starting point for iteration

Newton’s Method of Approximation by Iteration to find Zeroes on arbitrary function f(x):

Slope of Tangent to f(x) = [ f(xₙ) - 0 ] / [xₙ - xₙ₊₁] = f’(xₙ) ==> (xₙ - xₙ₊₁)f’(xₙ) = f(xₙ) ==> xₙ - xₙ₊₁ = f(xₙ) / f’(xₙ) ==> xₙ₊₁ = xₙ - f(xₙ) / f’(xₙ)

F(x) = ln(x) + 2 - (4/3)x ==> F’(x) = 1/x - 4/3

We have 2 “initial" values to inspect for points of intersection: (A) x₀ = 2.117, (B) x₀ = 0.17

Case (A) - x₀ = 2.117

x₁ = x₀ - F(x₀) / F’(x₀) ==> x₁ = 2.117 - [ ln(2.117) + 2 - (4/3)(2.117) ] / (1/2.117 - 4/3) = 2.117 - (-0.07267) / -0.861

x₁ = 2.033

x₂ = x₁ - F(x₁) / F’(x₁) = 2.2 - ( ln(2.033) + 2 - (4/3)(2.033) ) / (1/2.033 - 4/3) = 2.033 - (-0.00115) / (-.8414)

x₂ = 2.03163

x₃ = x₂ - F(x₂) / F’(x₂) = 2.03163 - ( ln(2.03163) + 2 - (4/3)(2.03163) ) / (1/2.03163 - 4/3) = 2.03163 - (-0.00000157) / (-.8411)

x₃ = 2.031628

==> We stop here at 5 decimal places precision ==> a_1 = 2.031628

Check:

f(x) = ln(x) + 2 ==> f(2.031628) = 2.708837

g(x) = (4/3)x ==> g(2.031628) = 2.708837

Case (B) - x₀ = 0.17

x₁ = x₀ - F(x₀) / F’(x₀) ==> x₁ = 0.17 - [ ln(0.17) + 2 - (4/3)(0.17) ] / (1/0.17 - 4/3) = 0.17 - (0.00138) / (4.55)

x₁ = 0.1697

x₂ = x₁ - F(x₁) / F’(x₁) = 2.2 - ( ln(0.1697) + 2 - (4/3)(0.1697) ) / (1/0.1697 - 4/3) = 0.1697 - (0.0000102) / (4.55942)

x₂ = 0.169698

x₃ = x₂ - F(x₂) / F’(x₂) = 0.169698 - ( ln(0.169698) + 2 - (4/3)(0.169698) ) / (1/0.169698 - 4/3) = 0.169698 - (0.000001107) / (4.55949)

x₃ = 0.169698

==> We stop here at 5 decimal places precision ==> a_0 = 0.169698

Check:

f(x) = ln(x) + 2 ==> f(0.169698) = .226265

g(x) = (4/3)x ==> g(0.169698) = .226264

Next: 2 Integrations in [ a_0 = 0.169698, a_1 = 2.031628 ]

Case I: Integrate F(x) on [a_0, a_1]

We want: Area Region R = A_R= ∫ₐ₀ᵃ¹ F(x) = ∫ₐ₀ᵃ¹ [ln(x) + 2 - (4/3)x] dx = (x*ln(x) - x) + 2x - (4/3)(x²/2) |ₐ₀ᵃ¹

A_R = x*ln(x) + x - (2/3)(x²) |ₐ₀ᵃ¹ = (0.720047) - (-0.150499) ==> A_R = 0.87055

Case II: Integrate G(y) on [b_0, b_1]

x = fᵧ(y) = e⁽ʸ ⁻ ²⁾ ==> y = ln(x) + 2

At x = a_0 = 0.169698 ==> y = b_0 = 0.22626

At x = a_1 = 2.031628 ==> y = b_1 = 2.70884

x = gᵧ(y) = (3/4)y ==> y = (4/3)x

At x = a_0 = 0.169698 ==> y = b_0 = 0.226264

At x = a_1 = 2.031628 ==> y = b_1 = 2.70884

Integrate G(y) = gᵧ(y) - fᵧ(y) ==> G(y) = (3/4)y - e⁽ʸ ⁻ ²⁾

Integrate G(y) on [b_0, b_1]

We want: Area Region R = A_R= ∫ᵦ₀ᵇ¹ G(y) = ∫ᵦ₀ᵇ¹ [(3/4)y - e⁽ʸ ⁻ ²⁾] dy = (3/4)(y²/2) - e⁽ʸ ⁻ ²⁾) |ᵦ₀ᵇ¹

A_R = (3/8)y² - e⁽ʸ ⁻ ²⁾ |ᵦ₀ᵇ¹ = (0.720047) - (-0.150499) ==> A_R = 0.87055

[Check] A_R = 0.87055 integrating both over ‘x’ and ‘y'

Observations: There is a lot of work above… but the question was only to write 2 Integrals !!

Well, you cannot write Definite Integrals unless you know the range of integration… unless we just want to get “hand-wavy” about it… sort of like… “then a miracle happens! And we have these numbers !"

So, what we did:

a. Study the functions, and neighborhoods of their intersections.

b. Drill down and obtain precise values for intersections, for range of Integration. I used the Newton-Raphson method… there are others, faster convergence, more setup, more messy.

c. Why all the work to get nearby “initial” values in order to “seed” the iterations ?

Because, without this, it is easy for the iteration to “pick up” a nearby root, other than the one we intended.

d. Express each curve in terms of ‘y’, in addition to ‘x’… flip the range of Integration to ‘y’.

e. I could have stopped here, with the 2 Integrals… but most of the work was already done.

f. I just kept going to complete both Integrations, and confirm the same area of Region R in both cases.

There is not always an advantage of 1 way over the other, on integration on ‘x’ vs. ‘y'. More often, context and information at hand will help determine this.

g. I did cheat a brief glance at Desmos to get a general idea at first, then confirmed details at the end... it is a good check.