Please help me with this math prob?

Question

Consider the region R in the 1st quadrant bounded above by the curve y=1/x between x=1 and x=5

a). find area of R

b). let x=b be a vertical line dividing R into 2 regions of equal area. With out doing any calculations, explain why b is less than 3

c). Use calculus to find the value of b in part b

d). find the volume of the solid with base R and whose cross sections cut by planes perpandicular to the x-axis are squares

Answer 1

a) area is ∫(1 to 5) 1/x dx
= ln 5 - ln 1
= ln 5.

b) Because y = 1/x is strictly decreasing on (1, 5), the area in the region (1, 3) will be more than the area in the region (3, 5) which has the same width.

c) ∫(1 to b) 1/x dx = ∫(b to 5) 1/x dx
=> ln b - ln 1 = ln 5 - ln b
=> 2 ln b = ln 5
=> b = e^((1/2) ln 5)
=> b = √5.

d). If the cross-sections are squares, each one will have area y^2. (If you sketch the solid it should look a bit like one of those old gramophone horns.) So the volume is
V = ∫(1 to 5) y^2 dx
= ∫(1 to 5) (1/x^2) dx
= -1/5 - (-1/1)
= 4/5.

Answer 2

1st you've to integrate 1/x = Ln X
2- evaluate (using X=1 as your inferior limit and x=5 as your superior limit)
LnX= Ln5-Ln1 =1.609-0=1.609 and this is the area

because 3 is the superior limit, and if you say that b divide R into 2 regions of equal area b can not be 3 because when X=3 you have an R area, and you are supossed to find the value in which x represents an area of R/2.

I don't remember the formula for the volume, the only thing that I can say is that you have to rotate the R area around the x axis, using dx.

I hope this could help you!!! :)

Answer 3

Interestingly the natural log of b (ln(b)) is sometimes defined as the area under 1/x from 1 to b. With this definition, ofcourse the answer to (a) is ln(5). You may also know that the antiderivative of 1/x is ln(x) + c, in which case you can use the fundamental theorem of calculus to see the same thing.

For (b), draw the graph and look at the line though x=3. You should see how to answer the problem.

For b and c, look in your calculus book.

Answer 4

a) The area is the integral dx/x: x|1-->5
b) Sorry on this one
c) Let A=A(x) be the area under the curve = integral in a)
then A/2 = integral dx/x: x|1-->b; solve for b knowing A from a)
d) Sorry here, I do not understand the question

Answer 5

Will want 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 will desire to drain the three litre and then flow the two litres from 5 litre to now empty 3litre fill the 5 litre and fill the three litre from it. this might take a million litre from the 5 litre leaving 4 litres