Integration problem regarding substitution. Posting below in a minute or two:

Question

mendicant_bias · Answer

"Use the substitution formula in Theorem 7 to evaluate the integral."

(Substitution formula is $$\int\limits_{a}^{b}f(g(x))*g'(x)dx = \int\limits_{g(a)}^{g(b)}f(u)du.$$
Where the integral given is:$$\int\limits_{-1}^{0}\sqrt{y+1} dy$$I'm not sure what i'm doing wrong, but here's how i' m trying to do it:

$$f(x) = \sqrt{u}$$$$u = (y+1)$$
$$g(-1) = (-1)+1 = 0 $$$$g(0) = (0)+1 = 1$$
$$\int\limits_{0}^{1}u ^{(1/2)}du = \int\limits_{0}^{1}\frac{ 2(y+1)^{3/2} }{ 3 }$$
$$\frac{ 2 }{ 3 }(1+1)^{3/2}-\frac{ 2 }{ 3 }(0+1)^{3/2} $$Which comes out to be something with two terms and a radical in the numerator in the first, it doesn't really matter, but it's definitely incoorect; the answer is 2/3.

anonymous · Answer

all good till
the very last step..
$$
\int_0^1u^{1/2}du={2\over3}[u^{3/2}]_0^1={2\over3}(1-0)=\mathbf{2\over3}
$$

anonymous · Answer

you do not convert it back to "y".

anonymous · Answer

but if you DO convert it, then your limits change back to the original limits

mendicant_bias · Answer

Just solved this during lunch with a friend, thanks, though. Well, you can convert it, right? It's just a choice of whether to do that or not, even though not converting it is much less tedious and much more sensible.