Little help with Logs/Integration Please 0_o (1 Viewer)

[chrome]

Q. Find the area between y = e^x and the x-axis for
log 1/2 < x < log4
\ \

yeh .. i'm a newb ~

 

[abdooooo!!!]

umm... is it 7.75/2 the answer?

just square it, times by pi, integrate it, then substitute the values.

oh wait its the area... lol... so just integrate it and sub in the values. 4 - 1/2 = 3.5

 

[McLake]

A = I{a->b} f(x)

Intergrate e^x
becomes [e^x]a->b
substitue limits (log 1/2 and log 4)
so (log4 - log1/2) = log8 units²

 

[wogboy]

substitue limits (log 1/2 and log 4)
so (log4 - log1/2) = log8 units2

Acutally when you sub in the limits log4 and log0.5, it becomes e^(log4) - e^(log0.5) = 4 - 0.5 = 3.5 (as abdoo posted) not log8.

It's too easy to make silly mistakes with such simple questions

 

[McLake]

Now I feel very stoopid!