Area between curve and axes (1 Viewer)

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Find the are between the curve [tex] y=\sqrt{1-x} [/tex] and the coordinate axes
 

NateTG

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Do you know integral calculus?
 

ShawnD

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I think it would be f(max) - f(min) where f(x) = (2/3)(1 - x)^(3/2)
 
You have to perform the integral
[tex]\int_{0}^{1}\sqrt{1-x}dx[/tex]
Try the substitution u=1-x
 
First find the domain and range that will give u limits of integration


Why you need a substitution

[tex]\int_{0}^{1}\sqrt{1-x}d(1-x)[/tex]
 
Originally posted by himanshu121

Why you need a substitution

[tex]\int_{0}^{1}\sqrt{1-x}d(1-x)[/tex]
I do believe this integral is equivalent to the one I posted. But how does your integral follow from the problem?
 
I was just shortening the step which are required for substitutions

Anyway i will be thinking that way too which u have asked
 
In order to do this problem, we usually take the following steps.
1. Sketch the curve [tex] y=\sqrt{1-x} [/tex] and find out what exactly you need to find.

2. Find the x-intercept(s) or y-intercept(s).

3. Write down a definite integral and solve the problem.

In this case, the x-intercept is 1, so you can find out the area by [tex]\int_{0}^{1}\sqrt{1-x}dx[/tex]

Originally posted by himanshu121
[tex]\int_{0}^{1}\sqrt{1-x}d(1-x)[/tex]
It should be
[tex]-\int_{0}^{1}\sqrt{1-x}d(1-x)[/tex]
 
Area is positive so in any case it is modulus
of the integral
 

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