Question

Could someone please verify if I got the right answer for the following integral (click to see).

Answer 1

\[\int\limits_{1}^{2}\frac{ x }{ \sqrt{4-x^2} }dx=3\]

Answer 2

Hmm I'm coming up with \(\large \sqrt{3}\), lemme check it on wolfram a sec.

Answer 3

Yah it's \(\large \sqrt{3}\). Where we running into trouble at? Did you apply a `u substitution`?

Answer 4

\[u=4-x^2 dx=\frac{ du }{ -2x }\]

Answer 5

\[\int\limits_{1}^{2}\frac{ x }{ \sqrt{u} }\frac{ du }{-2x }\]

Answer 6

\[=-\frac{ 1 }{ 2 }\frac{ u ^{1/2} }{ 1/2 }]_{1}^{2}\]

Answer 7

Oh ok I think I see the issue.
Did you at this point, back substitute, or did you evaluate it in u? Because I noticed that you didn't change your limits to u.

Answer 8

ok i found my mistake....when i brought (4-x^2) back it wasn't within a radical

Answer 9

that's why my answer is 3 and not sqrt(3)

Answer 10

Oh ok :) good catch.

Answer 11

thanks