Expression to an exponent: y = (x + 3)^4
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Harry_the_cat
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Yes.
So then this must be correct:
⌡ (x + 3)3dx = (x + 3)4 + C
........................4
If (x + 3)3 is expanded first and then integrated, it comes out:
x4 + 3x3 + 27x2 + 27x + C
4.............. 2
Why is it that if I put a value in for x, say 3, these are different?
(Note, that I had to use dots, ........., as place-holders as the text formatting here, for some reason, will not allow spaces to be used.)
⌡ (x + 3)3dx = (x + 3)4 + C
........................4
If (x + 3)3 is expanded first and then integrated, it comes out:
x4 + 3x3 + 27x2 + 27x + C
4.............. 2
Why is it that if I put a value in for x, say 3, these are different?
(Note, that I had to use dots, ........., as place-holders as the text formatting here, for some reason, will not allow spaces to be used.)
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Harry_the_cat
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Because the C's aren't equal.
\(\displaystyle \frac{(x+3)^4}{4} + C = \frac{x^4}{4}+3x^3+\frac{27x^2}{2}+27x +\frac{81}{4} +C\)
The second C "includes" the \(\displaystyle \frac{81}{4}\)
When you put in a value for x (say x=3) your answers will be different by \(\displaystyle \frac{81}{4}\).
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Harry_the_cat
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Consider \(\displaystyle y= x^2 + 6x + 7\) and \(\displaystyle y=x^2 +6x + 8\)
Graphically, the second graph is one unit higher than the first graph. At any x-value , their gradient (ie \(\displaystyle \frac{dy}{dx}\)) is the same.
In both cases \(\displaystyle \frac{dy}{dx}= 2x+6\)
So when you integrate \(\displaystyle 2x+6\), you get \(\displaystyle x^2+6x+c\). You can identify what c is if you have more information.