371 reputation
211
bio website
location
age
visits member for 1 year, 9 months
seen Oct 6 at 15:18

Studying CS. Into iOS apps at the moment.


Sep
26
awarded  Popular Question
Sep
24
awarded  Autobiographer
Aug
6
comment find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$
Ahh, I think I understand. The integral is from 0 to 1, the 1 comes from y=1?
Aug
6
comment find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$
Thank you! Using y seems easier, didn't realize I could do it like that. It seems like the y=1 just disappears, could you explain what happens with that for me?
Aug
6
comment find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$
I messed up... I forgot to get the antiderivative, I just plugged in -1 and 1 to the original equation...
Aug
6
revised find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$
added 83 characters in body
Aug
6
comment find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$
Yeah, I have. The book gives me a sketch. I got 3-sqrt(2). The book got 4/3.
Aug
6
asked find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$
Aug
5
accepted Evaluate the $\int_0^{1}{\cos(\frac{\pi t}2)}dt$
Aug
5
revised Evaluate the $\int_0^{1}{\cos(\frac{\pi t}2)}dt$
Possible answer
Aug
5
asked Evaluate the $\int_0^{1}{\cos(\frac{\pi t}2)}dt$
Aug
1
accepted Use the form of the definition of the integral to evaluate the integral (4-2x)
Aug
1
revised Use the form of the definition of the integral to evaluate the integral (4-2x)
[Edit removed during grace period]
Aug
1
asked Use the form of the definition of the integral to evaluate the integral (4-2x)
Jul
24
accepted Evaluate the limit: $\lim_{x\to \infty}$
Jul
24
comment Evaluate the limit: $\lim_{x\to \infty}$
@LucasVB Thanks, makes sense
Jul
24
accepted How to solve for $x$ for $\frac{1}2 x^{-1/2}- \frac14x^{-3/4}$
Jul
24
asked Evaluate the limit: $\lim_{x\to \infty}$
Jul
23
comment How to solve for $x$ for $\frac{1}2 x^{-1/2}- \frac14x^{-3/4}$
Awesome. I get x=1/16
Jul
23
asked How to solve for $x$ for $\frac{1}2 x^{-1/2}- \frac14x^{-3/4}$