find the area between the curves. y=log x y=x x=1 x=10
yay!! an easy one :)
let me draw a pic to see if we got the right interpretation
we integrate y=x between 1 and 10 then subtract y = log(x) between 1 and 10
\[\int\limits_{1}^{10} (x - \log(x)). dx\]
x^2 x ---- - ----- from [1,10] right? 2 log(x)
yes fro [1,10] and thanks for the graph!
x^2/2 = 100/2 - 1/2 = 99/2 99/2 - (10/log(10) - 1/log(1)) 99/2 - (10 - 1/0).... lol; lookslike I forgot how to int my log
thank you so much!
libraries closing on me so good luck with this one :)
can you please help me with my other questiones i posted :(
in about 10 minutes maybe; if they aint answered yet... gotta move to a new location :)
okay i will wait thanks!
log(x) = ln(x)/ln(10) 1/ln(10) is a constant so that gets put to the side for later int(ln(x)) = x(ln(x) - 1) xlnx - x ------- from 1,10 ln(10) 99/2 - [(10- 10/ln(10)) - (-1/ln(10))] is what i get
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