What is the real value of q for an area of 25 between two given functions?

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The problem and attempt at solution is attached in the word document. I think I have it right but I'm not sure about the upper and lower bound.
 

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How did you get the upper and lower bounds?
You should have set your two functions equal to each other and solve for x.
 
magicarpet512 said:
How did you get the upper and lower bounds?
You should have set your two functions equal to each other and solve for x.
I did. I got 6... I didn't get any other number so I assumed the lower bound was 2
 
setting your equations equal we have,
\sqrt{x - 2} = 8 - x
Solving for x should give us a quadratic equation to work with, which will have two roots.
Just apply the quadratic formula. You have one of the roots already, and it is not an upper bound.
 
Could someone help me with this question:
given two functions: H(x) and P(x),
H(x)=x^(2) and P(x) = 4-x^(2 )- q*x.
Note, the function P also has a parameter, q which is a real number.


Find the real value(s) of the parameter q such that the area of the region enclosed between these two functions is equal to 25.

I know that I have to H(x)=P(x) to get the intersecting points , but how could I get an answer if I don't have the coefficient q
 
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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