MHB Answer: Solving Algebra Problems: du=-30x^2dx to x^2 dx = -(1/30)du

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The transformation from du = -30x^2dx to x^2 dx = -(1/30)du involves dividing both sides by -30. This operation effectively isolates x^2 dx on one side of the equation. The result is then rearranged to express x^2 dx in terms of du. This algebraic manipulation clarifies the relationship between the variables. Understanding this step is crucial for solving similar algebraic problems.
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Hello,

Having a lapse in memory how was this 1) changed into 2) algebraically

1) du=-30x^2dx

2) x^2 dx = -(1/30)du thank you for any help.
 
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student231 said:
Hello,

Having a lapse in memory how was this 1) changed into 2) algebraically

1) du=-30x^2dx

2) x^2 dx = -(1/30)du thank you for any help.

It appears both sides were divided by -30, or equivalently, multiplied by $$-\frac{1}{30}$$, and then arranged as shown. Does that make sense?
 
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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