MHB How would you approach this integral?

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The integral is approached by substituting \( u = x - 3 \), leading to \( x = u + 3 \) and \( dx = du \). This transforms the integral into \( \int (u + 3)u^{1/2} du \), which simplifies to \( \int u^{3/2} + 3u^{1/2} du \). The resulting integral evaluates to \( \frac{2}{5}u^{5/2} + 2u^{3/2} + C \). Substituting back for \( u \) gives the final result as \( \frac{2}{5}(x - 3)^{5/2} + 2(x - 3)^{3/2} + C \). The method effectively demonstrates the use of substitution in solving integrals.
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$u=x-3$
 
Let u= x- 3. Then x= u+ 3 and dx= du so the given integral is equal to $\int (u+ 3)u^{1/2}du= \int u^{3/2}+ 3u^{1/2}= \frac{2}{5}u^{5/2}+ 2u^{3/2}+ C= \frac{2}{5}(x- 3)^{5/2}+ 2(x- 3)^{3/2}+ C$.
 

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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