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My final
- Thread starter Nick
- Start date
Nick said:Find all a greater than zero for which \(\displaystyle \int_{0}^{\infty}\frac{x}{a^{x^{2}}}\) converges and find it's value
\(\displaystyle \lim_{L\to \infty}\int_{0}^{L}\frac{x}{a^{x^{2}}}dx\)
Perform the integration and it becomes: \(\displaystyle \lim_{L\to \infty}\left[\frac{1}{2ln(a)}-\frac{1}{2a^{L^{2}}ln(a)}\right]\)
Now, what happens to the limit as \(\displaystyle L\to \infty\)?. What remains?. What values of 'a' is ln defined?.
Sketch the polar curve r=root theta * cos theta 0<=theta<= pi over 2
And find the area enclosed by the curve
Is that \(\displaystyle r=\sqrt{\theta cos\theta}, \;\ -\leq \theta \leq \frac{\pi}{2}\)?.
Nick said:For all a greater than zero except one?
For all \(\displaystyle a>1\)
No it is (root theta) times cos theta
\(\displaystyle r=\sqrt{\theta}cos\theta\)
\(\displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\theta\cdot cos^{2}(\theta)d\theta\)
I didn't get why it must be greater than 1 I'd get an answer if I had half which is less than one, can you please tell me why?
And about the other question will the graph be a petal in the first quadrant or will it be a circle or what exactly?
And thanks a lot for replying..
And about the other question will the graph be a petal in the first quadrant or will it be a circle or what exactly?
And thanks a lot for replying..