What is the area between the curve y=\sqrt{1-x} and the coordinate axes?

[tandoorichicken]

Find the are between the curve $$y=\sqrt{1-x}$$ and the coordinate axes

 

[NateTG]

Do you know integral calculus?

 

[ShawnD]

I think it would be f(max) - f(min) where f(x) = (2/3)(1 - x)^(3/2)

 

[StephenPrivitera]

You have to perform the integral
$$\int_{0}^{1}\sqrt{1-x}dx$$
Try the substitution u=1-x

 

[himanshu121]

First find the domain and range that will give u limits of integration


Why you need a substitution

$$\int_{0}^{1}\sqrt{1-x}d(1-x)$$

 

[StephenPrivitera]

Originally posted by himanshu121

Why you need a substitution

$$\int_{0}^{1}\sqrt{1-x}d(1-x)$$

I do believe this integral is equivalent to the one I posted. But how does your integral follow from the problem?

 

[himanshu121]

I was just shortening the step which are required for substitutions

Anyway i will be thinking that way too which u have asked

 

[KLscilevothma]

In order to do this problem, we usually take the following steps.
1. Sketch the curve $$y=\sqrt{1-x}$$ 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 $$\int_{0}^{1}\sqrt{1-x}dx$$

Originally posted by himanshu121
$$\int_{0}^{1}\sqrt{1-x}d(1-x)$$

It should be
$$-\int_{0}^{1}\sqrt{1-x}d(1-x)$$

 

[himanshu121]

Area is positive so in any case it is modulus
of the integral