Discover Common Mistakes: What Am I Doing Wrong?

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what am i doing wrong?

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You have the right value for the integral you show. If you are calculating area, then you shouldn't get a negative value, so your integral is set up wrong in that case.
 


If the problem was "Find the area between x= y^2- 4y and x= 2y- y^2 for y between 0 and 3" you should note that 2y- y^2> y^2- 4y for y between 0 and 3 so you have the subtraction backwards.
 
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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