How Did Community Support Help Improve My Calculus Grades?

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tony873004
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I just barely missed an A. I got a B in my calc I class :-p , thanks in large part to the people who answered my questions in this forum this semester, and those who answered other people's questions which I also found useful.

The first time I took Calc I, I got a C, but I feel I deserved an F. I understood nothing, and I wasn't ready for Calc II. This time I got a B and understood almost everything. It wasn't a lack of understanding. Only dumb mistakes on the final exam kept me from an A. So next semester, it's Calc II. Get ready for more questions :smile: .

Thanks again to everyone here who helped me with the tough homework problems this semester:wink:
 
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Congratulations, and I hope I helped you in some small way. :smile:
 
Yah, keep up the good work. Try your best, and you will get an A in your next calc course. :approve:
Good luck, :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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