I have to integrate the problem $e^{2x-1}$ and using $u$-substitution. I understand that I need to substitute $u=2x-1$ and $du=2dx$. My question is why I must have a $1/2$ outside of my integral before making the substitution. If you could explain each step in the integration so I can follow, where the step is necessary I would really appreciate it.
Integrating using U-Substitution
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calculus
1 Answers
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It's actually a matter of choice. Here's the way you're describing:
$$\int e^{(2x-1)} \, dx = \int e^{(2x-1)} \,\cdot 2\cdot\frac{1}{2}dx = \frac{1}{2} \int e^{(2x-1)} \cdot 2dx$$
We haven't changed the value of our integral, since multiplying by $2 \cdot \frac{1}{2}$ is just multiplying by 1. By pulling that $\frac{1}{2}$ out front, we now can clearly see the $2dx$ that we will substitute for $du$:
$$\frac{1}{2} \int e^{(2x-1)} \cdot 2dx = \frac{1}{2}\int e^u du$$
Alternatively, you could have made the substitutions $u = 2x-1$ and $\frac{du}{2} = dx$ from the start- and you would have ended up with the same integral.