Question

how do you find the area using integration?

Answer 1

Just put limits

Answer 2

limits? what about limits?..

Answer 3

integration is a means to add up an infinite number of non existing areas

Answer 4

you pretty much partition that area to be determined into rectangles; since rectangles are easy to find an area for.

then you make the rectangles skinnier and skinnier till they disappear.

then you add up all the disappeared areas and get the total

Answer 5

so you need to find the area of each rectangle?

Answer 6

yes; and depending on how accurate you want to be; you make the rectangles skinnier and skinnier

Answer 7

is it difficult to find the area?

Answer 8

it can be; it all depends on the functions you use for boundaries ...

Answer 9

polynomials are the easiest to integrate; more complicated functions such as sin(x^2)/5 can be trickier

Answer 10

how do you know the limits to be used?

Answer 11

the limits if they are to be used, correspond to the overall measurements of he area to be determined.

if you want to know the area for a box, your limits will be determined by its height width and depth

Answer 12

if you want to know the area between to intersecting curves; the limits will be the distance between the points where they meet

Answer 13

if you are given two equations and should find the area bounded by it?..what would be the first step?

Answer 14

its hard to put any "first step" to it. But ideally you would want to be able to know where the equations meet in order to establish the bounds of your limits

Answer 15

what if they meet in two points?

Answer 16

sometimes a first step involves reworking the equations to make them easier to deal with tho

Answer 17

find those 2 points :) usually by working out a system of equations like in algebra class

Answer 18

sometimes the firs step is to graph the equations so you get a feel for what they are doing

Answer 19

those two points corresponds to a coordinate what do i do with that?

Answer 20

you are trying to determine answers in an abstract way; this might be easier to deal with if you have a concrete example to work off of ...

Answer 21

ok. what is the form of the graph if the first equation of y is equal to the 2nd equation ?

Answer 22

ok. what is the form of the graph if the first equation of y is equal to the 2nd equation ?

Answer 23

are there steps? or several techniques to find the area using integration?

Answer 24

if the first equation is equal to the second equation for all xs and ys; then they are the same line or curve and the area between them is nothing .. if i read your question correctly

Answer 25

as far as techniques to integration; there are no set rules that will ever gaurentee you to find an answer

Answer 26

my teacher gave an example: find the area bounded by the curve y=x^2+5x and y=3-x^2. it was said that y=y..

Answer 27

there are some useful things you can "try" such as splitting up the inegrand into parts; trying to turn it into something easier to deal with that is

Answer 28

y=x^2+5x
y=3-x^2.

since these are polynomials they are easy to work with; helps build confidence to shatter later on :)

Answer 29

why is it y=y?

Answer 30

first step I would take is to determine the points where they meet

Answer 31

at the point where they meet; the y values and x values are the same; which is why they meet.

Answer 32

y = x^2+5x
y = 3-x^2

notice the when the equation turn out the same values for x and y they are at the same point on the graph; let me attach a picture

Answer 33

http://www.wolframalpha.com/input/?i=y+%3D+x^2%2B5x+and+y+%3D+3-x^2

Answer 34

from this you can see that there are 2 points that the curves intersect at; which means that the x and y values for each equation are identical at those points

Answer 35

to solve it with algebra, we have to recall techniques from systems of equations way back in algebra class

Answer 36

y = x^2+5x
y = 3-x^2

since y=y, there comes a point where each equation is equal to the other

x^2+5x = 3-x^2 ; equate the equations, and subtract one side
x^2 -3 -3+x^2 from the other
----------------
2x^2 +5x -3 = 0 .......... now since this is a normal quadratic, we
can solve for x

Answer 37

i don't understand why y=y.

Answer 38

2x^2 +5x -3 = 0 ; we can use quadratic formula for a simple
plug and play results

(1/4) (-5 +- sqrt(5^2 -4(2)(-3))) ; simplify
........

Answer 39

what do you understand "y" to be?

y = y means that there is some point or points at which these equation should equal each other ...

since y = A ,or y = B; then at some point A=B becasue they are both equal to y

Answer 40

spose we look at this as saying:

some car is traveling at 30 mph going east on a road; and another car is traveling at 50 mph going west on the same road; at the point that car = car we have a crash ....

Answer 41

i see that you are missing above, maybe you are having computer issues today :)

at any rate, we can relate the equation at the points where y = y; the points where they crash and get to what I typed above

x=\(\frac{1}{4} (-5 \pm \sqrt{5^2 -4(2)(-3)})\); simplify

=\(\frac{1}{4} (-5 \pm \sqrt{25 +24})\)

=\(\frac{1}{4} (-5 \pm \sqrt{49})\)

=\(\frac{1}{4} (-5 \pm 7)\)

=\(\frac{1}{4} (-12\ or\ 2)\)

=\(\frac{-12}{4} \ or \ \frac{2}{4}\)

x = -3 or 1/2

so our limits of integration will be from -3 to 1/2

Answer 42

in order to use these limits of integration we set up the problem like this:

\[\int_{-3}^{1/2}\ [f(x)-g(x)] \ dx\]

which can then be operated on in a few ways