A simple definite integral

Question

It has been a while since calculus and I am having a hard time remembering how to solve this problem.

$\int_{0}^{x} 9e^{-3x}dy$

I know that the answer is $9xe^{-3x}$, but I am not sure what the steps are.

Answer 1

$$\int _{ 0 }^{ x } 9e^{ -3x }dy=9e^{ -3x }\int _{ 0 }^{ x } dy=9e^{ -3x }\left( x-0 \right) =9e^{ -3x }x$$

Answer 2

You're integrating with respect to $y$, so as far as the integral is concerned, $x$ is just a constant. Therefore you can just pull $9e^{-3x}$ out of the integral like any old constant.

$$ \int_0^x 9e^{-3x} \, dy = 9e^{-3x} \int_0^x \, dy $$

Can you take it from here?

Answer 3

The key point is to notice that the integration is for $y$ and is not for $x$

$$\int_{0}^{x} 9e^{-3x}dy = 9e^{-3x} \int_0^xdy$$ $$=9e^{-3x}\cdot y \big|_0^x=9e^{-3x}(x-0)=9xe^{-3x}$$