Estimation is over or under estimate

dohsan
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Homework Statement


If y''=-y^2, is this approximationg an over-estimation or an underestimation of the exact area?


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The Attempt at a Solution

 
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Area of what? y''=(-y^2) is a differential equation. What does that have to do with an area?
 
I think this problem has to do with this equation... bc my teacher wrote it as a new problem and it confused me..

Write the Reimann sum that will approximate the area under the graph of the function y=f(x) that is continuous and postive over the interval [1,5] with n=8 using the right endpoint evaluation.
 
dohsan said:
I think this problem has to do with this equation... bc my teacher wrote it as a new problem and it confused me..

Write the Reimann sum that will approximate the area under the graph of the function y=f(x) that is continuous and postive over the interval [1,5] with n=8 using the right endpoint evaluation.

Ok, so to decide whether it is an overestimate or underestimate without actually solving the differential equation you need to at least decide whether y(x) is decreasing or increasing on [1,5]. Can you figure out how to do that?
 
Well, I believe you find y' which if f'(x), but not knowing what f is... it's quite confusing on how to find inc or dec on [1,5]. I just know if f'(x) > 0 .. it's inc.
 
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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