help with this math prob?

Question

the area of the region enclosed by the curve y = 1 /(x-1), the x axis, lines x = 3 and x = 4 is?

Answer 1

I would think that you would replace the 3 and 4 with the x-1(but im not too sure!!!)Like

y=1 y=1 y=1 y=1 So the answer would be
y=1/2 + y=1/3 then you

would say 2y=1+3y=1
then from there i forgot
____ ___ ____ ___
I helped a little bit-sorry!!!

x-1 3-1(=2) 2 4-1(=3)

Answer 2

the area is given by the integral (from a to b) of f(x) - g(x) dx

In this problem, f(x) = 1/(x-1) and g(x)=0, a=3, b=4
Then the area of the plane region is the integral (from 3 to 4) of [1/(x-1) - 0] dx
To find the integral of 1/(x-1)dx is to we let u=x-1, du=dx, so we can express it as the integral of 1/u du. This gives us lnu, which is ln(x-1) evaluated from 3-4.

Hence the area of the plane region is ln(4-1) - ln(3-1) which is just ln3 - ln2.

Answer 3

Looking at the graph, you want the area below 1/(x-1), above the x axis, to the right of x=3, and to the left of x=4.

Integrate 1/(x-1) from 3 to 4.

Int(1/(x-1)) = ln(x-1)

Evaluated from 3 to 4, ln(x-1) = ln(4-1) - ln(3-1) = ln(3) - ln(2) = .4055

Answer 4

Will desire an empty field Fill 5 litre Fill 3 litre from 5 litre........leaves 2 litres in 5 litre field Empty into spare field or drink Repeat and subsequently get or drink 4 litres or you are able to empty the three litre and then move both litres from 5 litre to now empty 3litre fill the 5 litre and fill the three litre from it. this may take a million litre from the 5 litre leaving 4 litres

Answer 5

use definate integrals ........... i think answer is ln(4/3)....... 'ln' is log base e

Answer 6

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