ertagon2
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Is the answer to this question correct?
View attachment 7783
View attachment 7783
Or, $(1-\cos(3x))\cdot\sin(3x) = \sin(3x) - \cos(3x)\cdot\sin(3x) = \sin(3x) - \dfrac{1}{2}\sin(6x)$ertagon2 said:Is the answer to this question correct?
MarkFL said:Well, let's see...let:
$$u=1-\cos(3x)\implies du=3\sin(3x)\,dx$$
So, the integral becomes:
$$I=\frac{1}{3}\int_0^1 u\,du=\frac{1}{6}(1^2-0^2)=\frac{1}{6}\quad\checkmark$$
You can think that if you like. However, it makes little or no difference whether we see the fraction version or the differential version - so long as the context is clear. The goal is to be clear and comprehensible. Notation is meant to express clearly what is wanted and to assist the communicator, not to become excessively cumbersome. Keep in mind, though that it may not be a good idea to invent new notation. There is already way too much. :-)ertagon2 said:Side question how is this [M]du=3sin(3x)dx[/M]allowed ? I thought that [M]du/dx is[/M] more of a notation than a fraction.
tkhunny said:Perhaps we can get some professional educators on here to give us a clue how tricky this is.
Q for Professional Educator: When a beginning calculus student first encounters the differential concept, is this usually an easy transition?
Learning the Derivative
If y = 2x^2, then dy/dx = 4x. I see, so dy/dx means the derivative of a function y, with respect to x.
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If y = 2x^2, then dy = 4x*dx. Wait, can you do that?