very curious (1 Viewer)

joeylike2hiphop

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hey smart ppl out dere i have a very curious question. how would we integrate Inx without using integration by parts, fanz ppl's
 

KFunk

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Hey dude, here's the gist:

y = log<sub>e</sub>x ---------> x = e<sup>y</sup>

I'll take that question from the cambridge 18 b) 28 as an example.
(Find &int; lnx dx between 1 and a)

At x=a , y=lna . Draw the perpendiculars from the x and y axis to (a,lna) and think in terms of a rectangle. The area of the rectangle is a.log<sub>e</sub>a and the area you want to get rid of is equal to the integral of e<sup>y</sup> between 0 and lna.

&int; e<sup>y</sup> dy between 0 and lna
= e<sup>(log<sub>e</sub>a)</sup> - e<sup>0</sup>
= a - 1

Hence the area of the area you're looking for is:
a.log<sub>e</sub>a - (a -1)
= a.log<sub>e</sub>a -a + 1

I hope that gives you the general idea.
 

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