# Antiderivatives!

#### Cervesa

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

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

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

you do the arithmetic to solve for $C$

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

• topsquark

#### topsquark

Forum Staff
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. -Dan

#### orcsmoocher

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

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

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

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