MHB Calculating Area Under a Curve: Is My Approach Correct?

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The discussion focuses on verifying the correctness of the approach to calculate the area under a curve using integrals. The first integral is correctly set up as the difference between the top and bottom functions, specifically $\int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$. The second problem involves the integral $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$, which is transformed using the substitution $u = x^2$. This substitution simplifies the integral to $\int_0^{\frac{\pi}{3}} sec^2(u)du$, confirming the setup is accurate. Overall, the calculations and transformations presented are valid for determining the area under the specified curves.
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Can you please check it for me that I have done it wrong or not ?
Thank you in advance.
 
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w/ respect to x, top function - bottom function

$\displaystyle \int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$

integral w/respect to y is ok and it can be simplified ...

$\displaystyle \dfrac{3}{2\ln{3}} \int_1^9 \ln{y} \, dy$

secant function integral is set up correctly ... antiderivative is rather easy to see if you recognize the chain rule
 
The second problem looks pretty standard- take x from 0 to $\sqrt{\frac{\pi}{3}}$ and, for each x, y from 0 to $2xsec^2(x^2)$.

The area is $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$.

Let $u= x^2$ so that $du= 2xdx$. When x= 0, u=0 and when x= $\sqrt{\frac{\pi}{3}}$, $u= \frac{\pi}{3}$.

The integral becomes $\int_0^{\frac{\pi}{3}} sec^2(u)du$.
 

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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