Is My Solution to the Improper Integral Correct?

[ku1005]

Hie, I have attempted the following inproper intergral, and I just wanted to check that I have answered it the correctly, if not, any tips would be great. I attached them as pics.

I attached my steps as pics, as I am not sure how to writ ein text on these forums.

My answer is thus = 0

Cheers

Rhys

PS if you can't view the pics for some reason?, um the Q is

Evaluate the improper intergral:

Intergal from 0 to 4 of

dx /[(4-x)^(1/3)]

 

[Office_Shredder]

The attachement is still pending, but here's a tip:

Take the derivative of your answer (not the constant for your final answer, but the function that the integral evaluated to, which you then plugged the endpoints into), and see if it matches up with what you're integrating

 

[HallsofIvy]

If you graph y= (4-x)^(-1/3) does it have some portion of its graph (between 0 and 4) below the x-axis? If not, the integral can't be 0!

Did you forget to evaluate the anti-derivative at x= 0?

 

[Gib Z]

I get -3\cdot \sqrt[3]{2}

 

[mjsd]

I get -3\cdot \sqrt[3]{2}

can't be negative, the function is above the x-axis from 0 to 4

edit: unless of course you meant anti-derivative evaluated at 0 only

 

[Gib Z]

Sorry I meant 3\cdot \sqrt[3]{2}

 

[ku1005]

I forgot to change my limits of integration (ie seen as I used a substitution method to evaluate the integral)

hence this is what I did

lower limit now = 4 and upper limit now = 4-a

hence the definite integral at the end becomes

lim a-> 4 from the left of -3(a)^(2/3) / 2

which gives me an answer of

[-3(4)^2/3]/2

which doesn't match your answer? so what am I doing wrong here??

thanks

 

[ku1005]

found it...was doing sum crazy stuff...really fundamentally stupid

now end up with POSITIVE

[3(4)^(2/3)]/2

which is equivalent to your answer I think

THANKS HEAPS!