integrating ln4x (1 Viewer)

[roosterman]

Can you guys help me out with this question?

Find the area bounded by the curve y = ln4x, the x-axis and x= 1/4 and x= 2.

I forgot how to integrate ln4x.

Cheers.

 

[followme]

i guess u hav to find the area of the curve bounded by the y-axis and subtract it from the rectangle. the area would be 2ln8-(7/4) i think.

 

[rama_v]

ln(4x) = ln 4 + ln x . then int (lnx) = xlnx - x (parts). go from there

Oops you do 2 unit. Try drawing e^x and find the corresponding area as suggested above.

 

[KYeh]

Assuming that lnx is already y²

f(x) = ln(4x)


∫f(x) dx = d/dx(4x)/4x + c
= 4/4x + c
= 1/x + c

then you can do the rest of the integration yourself.

Assuming if lnx is y, not y squared then,

f(x) = [ln(4x)] ²
= 2ln4x

then you can do the rest, hope it helps.

 

[acmilan]

Assuming that lnx is already y²

f(x) = ln(4x)


∫f(x) dx = d/dx(4x)/4x + c
= 4/4x + c
= 1/x + c

then you can do the rest of the integration yourself.

Assuming if lnx is y, not y squared then,

f(x) = [ln(4x)] ²
= 2ln4x

then you can do the rest, hope it helps.

The integral of ln(4x) isnt 1/x + c, the derivative of of it is 1/x.

What do you mean y², are you calculating volume?

 

[sinist4]

area is just 'y' ... i think you hav to draw it as ram a said

 

[KYeh]

Thanks for correcting me lol. I just saw what i wrote again and i was like "wtf are you on about!":burn:

 

[ThuanSUX]

I'll elaborate on what followme said:

Area required
= Area of rectangle - Area of curve and y-axis
= 2ln8 - int(x)dy (lim: 0->ln8)
= 2ln8 - int(1/4e^y)dy (lim: 0->ln8)
= 2ln8 - 1/4[8-1]
= 2ln8 - 7/4



*I got 1/4e^y by rearranging y=ln(4x) in terms of x
**0 and ln8 are boundaries in terms of y

 

[roosterman]

Just to let u guys know 2ln8 - 7/4 is the answer.

 

[brack777]

Are you saying the area of the rectangle is 2ln8? Shouldn't it be ln8 * (2 - 1/4) , there for answer = 7/4 * ln8 - 7/4