Question

what are the techniques in finding the area? using integration.

Answer 1

use y.dx and integrate

Answer 2

i think i can explain better with an example

Answer 3

find the area bounded by the curves y=x^2 + 5x and y= 3-x^2 let x=-2 and 0

Answer 4

A(x)=\[\int\limits_{b}^{a}(F(x)-G(x))dx\]

Answer 5

here is an example
cosider the graph
y=x
we want to calculate areA x FROM 0 to 1
we know it would be 1/2 (area of triangle)
usin integration
area=\[\int\limits_{0}^{1}x.dx\]
we get x^2/2
=1/2

Answer 6

http://en.wikipedia.org/wiki/Area_under_the_curve

Answer 7

i still dont understand

Answer 8

I think your x values are wrong. If not the functions you typed. Can you double check that it is the right interval. Also that the fxns are correct

Answer 9

This question is far too vague, u can write a book for the answer.

OP needs to be more specific.

Answer 10

A(x)=\[\int\limits_{a}^{b}(F(x)-G(x))dx\]

Typically, step 1 would be to calculate the x values and one can do that by setting the equations to equal each other for example: \[x^2+5x=3-x^2\]

Step 2: Solve for x values. This shows where both functions intercept and enclose a certain region. In the example problem you provided, solving for (x) would result in x=-3 and x=(1/2). (Looking at the graph will help). Typically, F

Step 3: Set up the equation. If I wanted to know the total area of the region enclosed by both functions I would use the definite integrate from x=-3 to x=(1/2) so b=(1/2) and a=(-3). HOWEVER, I think you wanted to find the area from the interval (0,1/2) so b=1/2 and a=0 thus making the equation:\\[[A(x)=\int\limits\limits_{0}^{(1/2)}((3-x^2)-(x^2+5x))dx\]\] (SETTING UP THIS EQUATION requires determining which function is on top of the enclosed region thus making it f(x) in the formula. so (upper-lower)

Step 4: Integrate and solve: \[A(x)=\int\limits_{0}^{1/2}((3-x^2)-(x^2+5x))dx=19/24\]

If I integrated using the right x values then the final answer should be A(X)=19/24