Diffucult integration question? (1 Viewer)

[p00_p00]

ummm can any1 do this question

y= e^x

Integrate between y axis and x =2???

I do it and the answer is wrong... the correct answer is 2ln2 -1 units

 

[p00_p00]

never mind got the answer....

 

[chanfky]

So how do you do it?

 

[FeniX]

you have to rearrange the equation to make X the subject

 

[chanfky]

rearrange the equation??
wouldn't it then become x=log y?
but u can't integrate a log

 

[Lazarus]

You don't have to re-arrange anything.

Simply integrate it between x=0 and x=2.

Unless he typed out the question wrong (which seems to be the case, looking at the answer he gave).

 

[p00_p00]

Yeah it should be between y axis and y=2

 

[Lazarus]

In that case, you need to use some good old ingenuity.

Integrating a log is outside the scope of the 2u (and 3u) course, so forget about doing that.

Notice that the graph of y = e^x is a reflection of y = ln(x) in the line y = x.

Therefore, the area under y = ln(x) between y = 1 and y = 2 is the same as the area under y = e^x between x = 1 and x = 2.

So, simply integrate y = e^x between x = 1 and x = 2.

At least, that's how I would have done it. Not sure why the answer was cited as 2ln2 - 1.

 

[Morgues]

Originally posted by Lazarus


At least, that's how I would have done it. Not sure why the answer was cited as 2ln2 - 1.

Laz wouldnt you do it this way.........ok you need the area encosed by e^x between y=2 and y=0

Now y=2 meets y=e^x at ln2........so you can construct a rectangle with width ln2 and height 2........so a rectangle with area of 2ln2

From this rectangle you need to deduct the area enclosed by e^x between x=ln2 and x=0 integral of e^x is e^x and this substituting of limits leads to e^ln2 - e^0 = 2-1 = 1

deducting this from area of rectangle is 2ln2 - 1

 

[Lazarus]

Oh, yeah.

I forgot about the other area on the LHS of the y-axis.

Listen to Morgues.