Find area of bounded region(s)?

Question

Can anyone help me on this problem?

Find the area of the region bounded by the graphs of the equations:

y=x^2lnx, y=0, x=1, x=e

I know the answer works out to be:

Area = 1/9(2e^3 +1)

but i'm lost on how to get there.

Any help is appreciated!

Answer 1

Set up an element of area dA and then integrate it from 1 to e to get the area.

dA = y*dx = (x^2)ln(x) dx

Use d(uv) = udv + vdu. Make udv the integral you want to find.

u = ln(x)  gives du = (1/x)dx

x^2ln(x)dx = x^2 u*dx which means x^2dx = dv and v = x^3/3

v*du = (x^3/3)(1/x)dx = (x^2/3)dx

d(uv) = u*dv + vdu

u*dv = d(uv) - vdu

u*dv = d{ln(x)x^3/3} - (x^2/3)dx

And each of these terms is easily integrated.

INTEGRAL(u*dv) = INTEGRAL(x^2ln(x)) = Area

Area = ln(x)x^3/3 - x^3/9 evaluated from x=1 to x=e

Area = [e^3/3 - e^3/9] - [-1/9]

Area = (1/9)(2e^3 + 1)

Answer 2

Firstly, find where the x-intercepts of y=x^2*ln(x) are, to ensure that negative area is made positive (assuming you want to find the total area, rather than the signed area)

0=x^2*ln(x)

x^2=0        

ln(x)=0

x-intercepts when x=0, x=1

For the area we are considering, the graph is positive (no x-intercepts exist between x=1 and x=e, and subbing in any value of x between 1 and 'e' yields a positive result)

To integrate the function, you will need to use integration by parts. For example, how would you work out the integral of y=ln(x)?

d/dx [x*ln(x)]= ln(x) + 1

Integrating both sides with respect to x;

x * ln(x) = ∫ [ ln(x) ] dx +  Ã¢ÂˆÂ« [1]dx

∫ [ ln(x) ] dx = [x * ln(x)] - x

The aim is to first differentiate another function which will give the original function you are looking at (it becomes easier with practice to recognize the function you need to differentiate). By the same token, we say;

d/dx {[(x^3)/3] ln(x)} = (x^2) * ln(x) + (x^2)/3

Integrating both sides with respect to x;

{[(x^3)/3] ln(x)}= ∫ [x^2 * ln(x)] dx + ∫ [(x^2)/3] dx

∫ [x^2 * ln(x)] dx= {[(x^3)/3] ln(x)} - (x^3)/9

Now, the upper limit and lower limit of this integral is 'e' and '1' respectively. Input these values into the equation (imagine the upper and lower limits are written... I don't know how to type them in...);

∫ [x^2 * ln(x)] dx

= {[(x^3)/3] * ln(x)} - (x^3)/9

=[(e^3)/3] * ln(e) - (e^3)/9 - [1/3] * ln(1) + 1/9

=[(e^3)/3] - (e^3)/9 + 1/9

= [ 2(e^3) + 1] / 9

=1/9*(2e^3 +1)

Answer 3

Make a graph and see how the structure looks like (sorry cant put the graph in this box)

Effectively its the area bounded by the x axis and the like y=x^2ln x from 1 to e.

So the area = definite integration with limits 1 to e for x^2 ln x

[Sorry i dont remember the integration formulae, so cant solve beyond the above step]