What is the area under the given equation using definite integration?

[Kahing]

Homework Statement

Find the area under the given equation using definite integration.
y = 4ln(3-x) from 0 to 2

The Attempt at a Solution

Integration from 0 to 2

4 (integration from 0 to 2) (3-x)
Let u = 3-x
du = -dx
-du = dx

-4(integration from 0 to 2) du

Idk where to go from there :/ sorry i do not know how to put the integral sign. I am currently new to calculus and i did not get it when my teacher explained it :/

 

[Dick]

Ok, so the integrand is now -4*ln(u)du. Not just -4du. What's the integral of ln(u)du?

 

[jhae2.718]

You can find $\int\!\ln(x)dx$ using integration by parts.

 

[Kahing]

Ok, so the integrand is now -4*ln(u)du. Not just -4du. What's the integral of ln(u)du?

thats what i don't understand. is 4ln(u), ln(u)^4 or 4*ln(u)

my attempt to integrate this problem would be u^n+1/n+1
5ln(3-x)/5?

 

[Dick]

thats what i don't understand. is 4ln(u), ln(u)^4 or 4*ln(u)

my attempt to integrate this problem would be u^n+1/n+1
5ln(3-x)/5?

That would be wrong. To integrate ln(x)dx pick du=dx and v=ln(x). So u=x and dv=(1/x)dx. Now do the parts thing. Integral of vdu=?

 

[Kahing]

That would be wrong. To integrate ln(x)dx pick du=dx and v=ln(x). So u=x and dv=(1/x)dx. Now do the parts thing. Integral of vdu=?

:( I really do not understand this. I don't understand integrating natural log itself. Is there a reference link you can show me ? I have googled it, but all there is is ln(1/x) dx never any ln(x)dx

 

[SammyS]

Do you know how to do integration by parts?

 

[Kahing]

Do you know how to do integration by parts?

actually i do not, that is 2 chapers ahead in my book :/

 

[Dick]

actually i do not, that is 2 chapers ahead in my book :/

Then try and guess the answer. You want to find a function whose derivative is log(x). Try taking the derivative of x*log(x). That's almost right. Can you add something to fix it up.

 

[hunt_mat]

Try differentiating $$y=x\ln x-x$$ and see what you get

 

[SammyS]

Look at the graph of y = 4ln(3-x).

The x intercept is x=2, and the y intercept is y = 4ln(3) .

Solving y = 4ln(3-x) for x gives x = 3 - e^y/4 .

From the graph of y = 4ln(3-x), it is apparent that the following integrals correspond to the same area.

$$\int_0^2{4\ln(3-x)}dx=\int_0^{\,4\ln(3)}{\left(3-e^{y/4}\right)}dy$$