Question

hello there!
quick question here...
if I have a definite integral, lower limit 1 and upper limit 4, when I substitute for 'u', u=sqrt(y), how do I know if u=1 or -1 or that u=2 or -2 on the new integral?

Answer 1

\[ u = \sqrt{y} \] is intrinsically positive. The radical sign implies the positive square root.

Answer 2

What is the integrand of the original integral?

Answer 3

Hello there! Thank you so much for this... the original integrand is:

\[\int\limits_{1}^{4} (x-1)/\sqrt{x}\]

Answer 4

\[\int\limits_{1}^{4}(x-1)/\sqrt{x}=\int\limits_{1}^{4}(x/\sqrt{x}-1/\sqrt{x})=\int\limits_{1}^{4}(x^{1/2}-x^{-1/2})=[2/3*x^{3/2}-2x^{1/2}]_{1}^{4}\]
\[=2/3*8-4-(2/3-2)=14/3-2=8/3\]

Answer 5

thank you very much! yes, I got this too... I just was not sure about what to do with the substitution ( I did it a bit differently, I used a substitution: u= \[\sqrt{x}\]

Answer 6

happy studies!
I did an exam today.. but not sure I did too well! I hope I get a rewrite at least :/