Integration involving Square Roots (1 Viewer)

[_ShiFTy_]

E.g.
Integrate
√(9-x^2) between x=0 and x=3

I know you can use the area of a circle thing but is there a way to do this another method? If its possible but only in 4U, please say so cos i also do 4U

 

[pLuvia]

I think you use log for this question, then integrate from there

 

[香港!]

E.g.
Integrate
√(9-x^2) between x=0 and x=3

I know you can use the area of a circle thing but is there a way to do this another method? If its possible but only in 4U, please say so cos i also do 4U

I=int. √(9-x^2) dx
let u=√(9-x^2) and dv\dx=1
du\dx=(1\2)(9-x²)^(-1\2) and v=x
I=x√(9-x^2)-(1\2) int. x\√(9-x²) dx

let u=9-x² to integrate the x\√(9-x²) then ur done

 

[Riviet]

There is also a method that i've seen by substituting trig into the integral.

 

[香港!]

^ yea put x=3cos@

 

[insert-username]

Integrate √(9-x²) between x=0 and x=3

There is a much, much easier way to do this type of question - the area of a circle, like you said. Don't bother with integration if you can use easier methods. This function is a semicircle above the x-axis with radius 3. Thus, its integral from 0 to 3 is equal to half the area of the semicircle (9pi/2), or one quarter of the area of a circle with radius 3 (9pi), i.e. 9pi/4.


I_F

 

[Riviet]

But the question asked us to "integrate the function". Therefore we need to use the rules of integration to integrate it, not any other formulae that we know of, such as volume of sphere, which would be much easier than if we were to rotate the function about the x-axis. However, we can use the circle formula if the question asked us to "find the area" under the curve or between the curve and x-axis, since it didn't directly ask for integral methods.

 

[icycloud]

Correct me if I'm wrong, but weren't the area and volume formulae initially developed using calculus anyway?

 

[insert-username]

But the "integral" of a function is the total area of rectangles drawn from point a to point b under the function as the width of those rectangles goes to 0. Hence, the circle method is perfectly valid because it finds the exact area of each of the little rectangles under the function from 0 to 3 - the integral.


I_F

 

[icycloud]

But the "integral" of a function is the total area of rectangles drawn from point a to point b under the function as the width of those rectangles goes to 0. Hence, the circle method is perfectly valid because it finds the exact area of each of the little rectangles under the function from 0 to 3 - the integral.


I_F

Yeh but that's using a derived formula from the process of integration itself. So it's like using a differentiation 'rule' when it says to derive by first principle. This is just my take on the issue.

 

[insert-username]

That post was directed to Riviet, not you. Sorry.


I_F

 

[Riviet]

Correct me if I'm wrong, but weren't the area and volume formulae initially developed using calculus anyway?

That is correct icycloud, i have proved the volume of a sphere formula before.

But the "integral" of a function is the total area of rectangles drawn from point a to point b under the function as the width of those rectangles goes to 0. Hence, the circle method is perfectly valid because it finds the exact area of each of the little rectangles under the function from 0 to 3 - the integral.

. . . . . . f a i r e n o u g h t h e n.

Just be careful in an exam situation of which method you do though.

 

[pLuvia]

That is correct icycloud, i have proved the volume of a sphere formula before.



. . . . . . f a i r e n o u g h t h e n.

Just be careful in an exam situation of which method you do though.

Me too ! with tutor and school