Please answer this calculus question?

Question

Consider the region R in the first quadrant under the curve y=(2*ln(x))/x^2. 

 

a) Write the area of R as an improper integral.

b) Express the integral in part (a) as a limit of a definite integral. 

c) Find the area of R. 

 

Please show all your work. 

Answer 1

Consider the region R in the first quadrant under the curve f(x) = 2*ln(x)/(x^2).

~ First Quadrant ==> x ≥ 0, y ≥ 0.

y = 2*ln(x)/(x^2) ≥ 0

Multiply both sides by x^2/2 (which is positive):

ln(x) ≥ 0

e^ln(x) ≥ e^0

x ≥ 1; so limits of integration are {1,∞}.

a) Write the area of R as an improper integral.

R = 2 ∫{1,∞}((ln(x)/(x^2)) dx)

b) Express the integral in part (a) as a limit of a definite integral.

R = 2 limit{t → ∞}(∫{1,t}((ln(x)/(x^2)) dx))

c) Find the area of R.

u = ln(x), u(1) = 0, u(t) = ln(t)
e^u = x
du = dx/x
I = ∫{1,t}((ln(x)/(x^2)) dx)
= ∫{0,ln(t)}((u/(e^u)) du)

redefine u as x (dummy variable of integration, not same x as above)

I = ∫{0,ln(t)}((x e^(-x) ) dx)

d(uv) = (du)v + u(dv)
(du)v = d(uv) – u(dv)
v = x, du = e^(-x) dx
dv = dx, u = -e^(-x)

I = ∫{0,ln(t)}(v du)
= [uv - ∫(u dv)]{0,ln(t)}
= [-x e^(-x) - ∫(-e^(-x) dx)]{0,ln(t)}
= [-x e^(-x) - e^(-x)]{0,ln(t)}
= [-e^(-x)(x + 1)]{0,ln(t)}
= -e^(-ln(t))(ln(t) + 1) + e^(0)(0 + 1)
= -(1/t)(ln(t) + 1) + 1
= -ln(t)/t - 1/t + 1

R = 2 limit{t → ∞}[-ln(t)/t - 1/t + 1]

t → ∞ ==> ln(t)/t → ∞/∞
Indeterminate form, so can use L’Hospital’s Rule.
t → ∞ ==> ln(t)/t → (ln(t))’/(t)’ → (1/t)/1 → 0

R = 2 [0 - 0 + 1] = 2

GeoGebra check using R ≈ Integral[f, 1, 10^6]: http://www.wyzant.com/resources/files/267500/improper_integral