Hey Chaim.
The answer for the 3 and 1/3 is that du = 3dx.
This means that 1/3 * 3 dx = dx = 1/3 * du and hence if you are integrating with respect to u then you need a 1/3 term to balance everything out.
Hi, I’m a bit confused on how to solve a problem like this.
My friend helped me a bit, but didn’t explain very clearly, so I was wondering if you guys could help me understand this.
_{0}∫^{1/6} 1/√(1-9x^{2})dx
So I was told that u=3x, and du=3
So it would become _{0}∫^{1/6} 1/√(1-(3x)^{2}dx
Now for u-substitution.
1/√(1-u^{2})du
I get [arcsin(3x)]^{1/6}_{0}
Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:
(1/3)_{0}∫^{1/6} (1/√(1-(3x)^{2}) (3)dx
So yeah… there’s a 3 outside…
And there’s a 1/3 in front of the integration.
Can someone explain to me why this was done?
Also, can someone explained why u=3x, instead of something like u=9x?
Thanks J
Hi, I’m a bit confused on how to solve a problem like this.
My friend helped me a bit, but didn’t explain very clearly, so I was wondering if you guys could help me understand this.
_{0}∫^{1/6} 1/√(1-9x^{2})dx
So I was told that u=3x, and du=3
So it would become _{0}∫^{1/6} 1/√(1-(3x)^{2}dx
Now for u-substitution.
1/√(1-u^{2})du
I get [arcsin(3x)]^{1/6}_{0}
Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:
(1/3)_{0}∫^{1/6} (1/√(1-(3x)^{2}) (3)dx
So yeah… there’s a 3 outside…
And there’s a 1/3 in front of the integration.
Can someone explain to me why this was done?
Also, can someone explained why u=3x, instead of something like u=9x?
Thanks J