Antiderivatives!

Nov 2019
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0
Depths of Hell, babeeey
(Ignore the 3rd part for now, I haven't tried it yet.)
2054a968d3a1fb4e4f95861022d4cce4.png
I got the answer listed above, and inserted 1 into the equation and got -7 back. I'm not sure how it's wrong?
Here's my work.
 

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Dec 2014
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$g(x) = 3x^{-3}-8x^{-7}$

$G(x) = 3 \cdot \dfrac{x^{-2}}{-2} -8 \cdot \dfrac{x^{-6}}{-6} + C = -\dfrac{3}{2x^2} + \dfrac{4}{3x^6} +C$

$G(1)=-7=-\dfrac{3}{2}+\dfrac{4}{3} + C$

you do the arithmetic to solve for $C$

use the same process to solve for $H(x)$
 
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topsquark

Forum Staff
Jan 2006
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You were trying to use the power rule that says \(\displaystyle \int x^n ~ dx = \dfrac{1}{n + 1} x^{n + 1} + C \).

Note that \(\displaystyle \int \dfrac{1}{x^3} ~ dx = \int x^{-3} ~ dx = \dfrac{1}{(-3) + 1} x^{(-3) + 1} + C = \dfrac{1}{-2} x^{-2} + C = -\dfrac{1}{2 x^2} + C\)

You fell into the trap of trying to say -3 + 1 = -4 because "-4 is one bigger than -3". (4 is bigger than 3, right?) This can be very hard to catch and usually occurs when you are working too fast or on not enough sleep. šŸ˜“

-Dan
 
Nov 2019
23
0
Depths of Hell, babeeey
$g(x) = 3x^{-3}-8x^{-7}$

$G(x) = 3 \cdot \dfrac{x^{-2}}{-2} -8 \cdot \dfrac{x^{-6}}{-6} + C = -\dfrac{3}{2x^2} + \dfrac{4}{3x^6} +C$

$G(1)=-7=-\dfrac{3}{2}+\dfrac{4}{3} + C$

you do the arithmetic to solve for $C$

use the same process to solve for $H(x)$
Got it! Thank you!

You were trying to use the power rule that says \(\displaystyle \int x^n ~ dx = \dfrac{1}{n + 1} x^{n + 1} + C \).

Note that \(\displaystyle \int \dfrac{1}{x^3} ~ dx = \int x^{-3} ~ dx = \dfrac{1}{(-3) + 1} x^{(-3) + 1} + C = \dfrac{1}{-2} x^{-2} + C = -\dfrac{1}{2 x^2} + C\)

You fell into the trap of trying to say -3 + 1 = -4 because "-4 is one bigger than -3". (4 is bigger than 3, right?) This can be very hard to catch and usually occurs when you are working too fast or on not enough sleep. šŸ˜“

-Dan
Ha, makes sense. As soon as I finished Set 8 I immediately jumped to this one.