# 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

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

note that $\frac{1}{y^a}=y^{-a}$.

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

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

Does this make sense?

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

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

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

Does this make sense?

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