Question

Hi a question on intergration:

By considering the area under the curve y=√x between x=0 and x=81 as sum of strips of width 1 unit show that.

∫(integrating from 0 to 81) √xdx<√1+√2+√3+...√81.

Answer 1

any idea how i start the question

Answer 2

I would depend on if your using right or left end points

Answer 3

starting at x=0, then using strips one unit wide would actually give you a lesser estimate

Answer 4

ok would i need to go up to 81 or would i recognise the pattern after a couple of terms? and can you help me start of

Answer 5

sure no problem ...

Answer 6

are you familiar with the shape of the graph sqrt(x)?

Answer 7

yes its only positive y values

Answer 8

and the way it curves is also important

Answer 9

what we are doing here is making rectangles 1 unit wide, that have a height of sqrt(x) for all interger values of x between 0 and 81

Answer 10

then we are summing the areas of these rectangles.

Answer 11

area of a rectangle = width times height

Answer 12

so our equation from our strips will be A = Width of rectangle X Height of rectangle

Answer 13

o ok so the width is dx and x is height

Answer 14

Sqrt(x) is the height

Answer 15

picture a rectangle under the curve of the graph that only touchs the curve at one point, that point is our height

Answer 16

i should have seen that from initial expression

Answer 17

you could have..

Answer 18

ok so because our dx=1, or our width is one we have A=sqrt(x)

Answer 19

The integral estimate on [0,81] then becomes sqrt(1) + sqrt(2) + ... + sqrt(81)

Answer 20

this is what our estimate would roughly look like, notice the negative space, this is why our estimate would be less then the actual integral