MHB Can You Help Me Integrate $\sin(u) + u^6$ Correctly?

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The discussion focuses on integrating the expression $\sin(u) + u^6$. The initial confusion arose from misplacing parentheses, leading to the incorrect interpretation of the function as $\sin(u + u^6}$. Clarification was provided that the correct expression is simply $\sin(u) + u^6$, which simplifies the integration process. The participant expressed gratitude after understanding the correct approach. Proper placement of parentheses is crucial for accurate mathematical interpretation and integration.
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Can you please help me ?
I have tried to do it many times but still got the wrong answer.
Thank you in advance.
 
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Make the substitution $t = \sin x$.
 
Opalg said:
Make the substitution $t = \sin x$.
Untitled 5.png
I am stuck here
Can you please help me ?
 
Get the parentheses in the right place! It's not $\sin(u+u^6)$, but $\sin(u) + u^6$, which is much easier to integrate.
 
Opalg said:
Get the parentheses in the right place! It's not $\sin(u+u^6)$, but $\sin(u) + u^6$, which is much easier to integrate.
Thank you very much
I understand 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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