MHB Finding Area between 2 functions

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To find the area between the curves y = x^2 and y = 2/(x^2 + 1), the points of intersection are identified as -1 and 1, confirming symmetry. The area can be calculated using the integral 2∫(0 to 1) (2/(x^2 + 1) - x^2) dx. A standard integral, ∫(dx/(x^2 + 1)) = arctan(x), is suggested to assist in solving the problem. The discussion emphasizes the need for proper integration techniques to arrive at the correct area.
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Hi,

I have this problem to find the area between 2 curves:

$y = x^2$

and

$y = \frac{2}{x^2 +1}$

I found that the points of intersection are -1 and 1 and it is symmetrical.

I get
$2\int_{0}^{1} \ \frac{1}{x^2 + 1} - x^2 dx$which I am unable to solve. I have tried u-substitution but I end up getting mixed up.

Thanks
 
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There's a standard integral involved:
$$\int \frac{dx}{x^2+1} = \arctan(x)$$

I think you can figure out the answer right now ;)
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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