# Thread: need help with "simple" integration

1. ## need help with "simple" integration

Hey everybody, first time using the forum because im taking cal II this semester and Im sure ill need help : )

I dont know how to integrate when an exponent is in the denominator. For ex:
4/y^2

my book does this:

4/y^2 = 4(-1/y) when integrated and i dont know how they did it, im sure its a combination of some algebraic trick that im just not catching.

For reference the book is: Thomas Calculus early transcendentals pg. 431 ex. 7

thanks for any help you can offer

2. Originally Posted by cknicker
Hey everybody, first time using the forum because im taking cal II this semester and Im sure ill need help : )

I dont know how to integrate when an exponent is in the denominator. For ex:
4/y^2

my book does this:

4/y^2 = 4(-1/y) when integrated and i dont know how they did it, im sure its a combination of some algebraic trick that im just not catching.

For reference the book is: Thomas Calculus early transcendentals pg. 431 ex. 7

thanks for any help you can offer
note that $\displaystyle \frac{1}{y^a}=y^{-a}$.

So in your case $\displaystyle \frac{4}{y^2}=4y^{-2}$.

When you integrate, you apply power rule to get $\displaystyle \int 4y^{-2}\,dy=\frac{4y^{-2+1}}{-2+1}+C=-4y^{-1}+C=-\frac{4}{y}+C$

Does this make sense?

3. Originally Posted by Chris L T521
note that $\displaystyle \frac{1}{y^a}=y^{-a}$.

So in your case $\displaystyle \frac{4}{y^2}=4y^{-2}$.

When you integrate, you apply power rule to get $\displaystyle \int 4y^{-2}\,dy=\frac{4y^{-2+1}}{-2+1}+C=-4y^{-1}+C=-\frac{4}{y}+C$

Does this make sense?

yes it does now, thank you very much for your help.