Why can't 4*sqr(1-x^2) be integrated from 0 to 1 to get an exact value?

Question

How?

Answer 1

Use a trig substitution.
Draw a right triangle with the right angle at the bottom right.
label the bottom left angle as theta. label the bottom side as x. Label the hypoteneuse as 1.
By the pythagorean theorem, the right side will be sqr(1-x^2)
Based on this triangle:
sin(theta) = sqr(1-x^2)
cos(theta) = x
Take the derivitive of this to get
dx = -sin(theta) dtheta

Now you can substitute these in your original problem to get
4*sqr(1-x^2)dx = 4*sin(theta) * -cos(theta) dtheta

At this point you can continue by doing it a couple different ways.
Since you changed variables, you need to change the limits of integration, using the equation cos(theta) = x, or in other words
theta = arccos(x). Then integrate using the new limits.

Another way is to not change the limits of integration. Integrate it. Then, before using the limits of integration, substitue back in x based on the triangle we used above. for instance if you have sin(theta) , replace it with sqrt(1-x^2). then use your limits of integration that you started with to finish it.

Answer 2

ok, at the beginning, we ought to discover dt in terms of dx to alter the vital. we additionally will ought to alter the limitations besides. So, at the beginning making use of: t = sqrt(x+a million) supplies dt = a million/2(x+a million)^(-a million/2)dx Re-arranging supplies dx = 2(x+a million)^(a million/2) dt. Now placed this into our unique equation, changing the dx, and the different x with (t^2-a million) supplies: vital [(t^2-a million).2dt] Multiplying out supplies: vital [2t^2 - 2 dt.] This integrates to 2/3t^3 - 2t. we ought to alter the limitations making use of our substitution besides the fact that which provides 0 -->a million, 3 --> 2 (making use of t = sqrt(x+a million). So making use of those limits supplies: (sixteen/3-4) - (2/3-2) which provides: 8/3.

Answer 3

How? Trigonometric substitution!

Let x=sin(y)
then
dx/dy = cos(y)
or
dx = cos(y)dy

Int (0 to 1) 4*sqrt(1-x^2) dx
= Int (0 to pi/2) 4*sqrt(1-sin(y)^2)cos(y)dy
Recall cos(y)^2 = 1-sin(y)^2
= 4*Int(0 to pi/2) cos(y)^2dy
= 4*[y/2 + sin(2y)//4] <--evaluated between (0,pi/2)
= 4 *(pi/4) = pi

Answer 4

not a clue

Answer 5

you got me

Answer 6

I have no Idea.

Answer 7

Uhm. I think it can be.