Simple Integration Question (1 Viewer)

[the-derivative]

This may probably seem like a dead easy question for most people, but I can't seem to get it.

Find the area of the region bounded by the curve 3x/(x^-1), the x-axis and the line x = 1/2.

This is what I've done so far:

Let u = x^2 - 1
du = 2x dx
du/2= x dx

also changing the boarders:

when x = 1/2, u = -3/4
x = 0, u = -1

therefore subbing the new values in I get

the integration between -1 and -3/4 of 3/2[du/u]

which equals to 3/2[ln u] between -1 and -3/4.

The main trouble is when I sub in the boarders because then I'll get ln of a negative number.

Thank you in advance for your help.

 

[ianc]

This may probably seem like a dead easy question for most people, but I can't seem to get it.

Find the area of the region bounded by the curve 3x/(x^-1), the x-axis and the line x = 1/2.

This is what I've done so far:

Let u = x^2 - 1
du = 2x dx
du/2= x dx

also changing the boarders:

when x = 1/2, u = -3/4
x = 0, u = -1

therefore subbing the new values in I get

the integration between -1 and -3/4 of 3/2[du/u]

which equals to 3/2[ln u] between -1 and -3/4.

The main trouble is when I sub in the boarders because then I'll get ln of a negative number.

Thank you in advance for your help.

hi there!!!

you've done all the hard work, just keep plowing on and don't worry about the negative numbers...

= 3/2 [ln u] evaluated between u = -1 and u= -3/4

= 3/2 [ln(-3/4)] - 3/2 [ln(-1)]

now use the log law ln(x)-ln(y) = ln(x/y) and you'll find that you now have a real number.

isn't maths cool?

 

[the-derivative]

hi there!!!

you've done all the hard work, just keep plowing on and don't worry about the negative numbers...

= 3/2 [ln u] evaluated between u = -1 and u= -3/4

= 3/2 [ln(-3/4)] - 3/2 [ln(-1)]

now use the log law ln(x)-ln(y) = ln(x/y) and you'll find that you now have a real number.

isn't maths cool?

HAHA thanks heaps man.

Yeah maths is awesome xD

 

[tommykins]

This may probably seem like a dead easy question for most people, but I can't seem to get it.

Find the area of the region bounded by the curve 3x/(x^-1), the x-axis and the line x = 1/2.

This is what I've done so far:

Let u = x^2 - 1
du = 2x dx
du/2= x dx

also changing the boarders:

when x = 1/2, u = -3/4
x = 0, u = -1

therefore subbing the new values in I get

the integration between -1 and -3/4 of 3/2[du/u]

which equals to 3/2[ln u] between -1 and -3/4.

The main trouble is when I sub in the boarders because then I'll get ln of a negative number.

Thank you in advance for your help.

quicker way -

int. 3x/(x^2-1) dx = 3/2 int. 2x/x^2-1 = 3/2 ln (x^2-1)