Train Wreck Integral Shortcut...

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SUMMARY

The discussion centers on the evaluation of the definite integral $$\int_1^2 x^3 dx$$ and the confusion surrounding the application of integral properties. The correct evaluation yields $$\left[\frac{x^4}{4}\right]_1^2 = \frac{15}{4}$$, while the incorrect assumption that $$\left[\frac{x^4}{4}\right]_1^2$$ equals $$\left[x^2\right]_1^2\left[\frac{x^2}{4}\right]_1^2$$ leads to an erroneous result of 9. Participants emphasize the importance of understanding integral properties and suggest using trigonometric identities for simplifying integrals involving trigonometric functions.

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aronclark1017
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TL;DR
A tempting shortcut in long paths that leads to certain train wreck
Although this case is seemingly reasonable.
I(1,2) x^3 dx
[x^4/4](1,2)
[x^2](1,2)[x^2/4](1,2)=9/4

I(1,2) x^3 dx
[x^4/4](1,2)=15/4
 
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It helps if you use LaTeX. Is your question that if $$\int_1^2x^3dx=\left[\frac{x^4}4\right]_1^2$$then why is $$\left[\frac{x^4}4\right]_1^2\neq\left[x^2\right]_1^2\left[\frac{x^2}4\right]_1^2$$Because if so, write out the square brackets explicitly and you'll see why.
 
##(ab)|_1^2\neq a|_1^2\,b|_1^2##
 
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aronclark1017 said:
Although this case is seemingly reasonable.
Not at all.
I'll simplify your problem slightly by getting rid of the fractions.
##\left. x^4\right|_1^2 = 2^4 - 1^4 = 16 - 1 = 15##

##\left. x^2\right|_1^2 \cdot \left. x^2\right|_1^2 = (2^2 - 1^2)(2^2 - 1 ^2) = (4 - 1)(4 - 1) = 3\cdot 3 = 9##
 
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Order in the court is cooks law MF trust is..🤣🤣🤣
 
aronclark1017 said:
Order in the court is cooks law MF trust is..🤣🤣🤣
@aronclark1017, you don't do yourself any favors by posting gibberish such as the above.
 
Yea, Its just that I don't recall any such law in the text. Although I'm sure there is a name for this one. Especially since it seemingly would greatly simplify some integrals if it did work. For example

x^3 cos^2(t) +x^3 cos(t) sin(t) + x^2 sin(t)

Sure would be nice to factor out that x^2 before solving the definite integral on x. A veteran expert might just say well if the bound is on 0,2pi then the last two terms cancel anyway and bypass the law entirely and get his PHD before ever realizing such law ever existed. Either that or just erase the entire problem and start from scratch until the correct solution is derived, however it happened. It's why just scratching at problems by hand like playing music is not good especially at high levels. There is a need to just read things. In fact, it's the law. Because the hand is not a copulator, is simply a pointer to the problem which must first be read hands free.
 
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aronclark1017 said:
Yea, Its just that I don't recall any such law in the text. Although I'm sure there is a name for this one.
I'm pretty sure that there is no such law, and definitely not one named Cook's Law.
aronclark1017 said:
x^3 cos^2(t) +x^3 cos(t) sin(t) + x^2 sin(t)
One can start with the first two terms by using trig identities. For the first, the identity ##\cos^2(t) = \frac 1 2 (\cos(2t) + 1)## can be used. For the second term, the identity ##\sin(t)\cos(t) = \frac 1 2 \sin(2t)## can be used. Once these are done, then the usual approach for all three terms is integration by parts or you can look in a table of integrals.
 

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