Floyd Johnston
04/30/2023 · Elementary School
Determine the following. \( \int 2 e^{-0.5 \mathrm{x}} \mathrm{dx} \) \( \int 2 e^{-0.5 \mathrm{x}} \mathrm{dx}=\square \) (Type an exact answer.)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To solve the integral \( \int 2 e^{-0.5 x} \, dx \), we can use the method of integration for exponential functions.
1. **Identify the integral**:
\[
\int 2 e^{-0.5 x} \, dx
\]
2. **Factor out the constant**:
\[
= 2 \int e^{-0.5 x} \, dx
\]
3. **Integrate the exponential function**:
The integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} + C \), where \( a \) is a constant. Here, \( a = -0.5 \).
Therefore,
\[
\int e^{-0.5 x} \, dx = \frac{1}{-0.5} e^{-0.5 x} + C = -2 e^{-0.5 x} + C
\]
4. **Substituting back into the integral**:
\[
2 \int e^{-0.5 x} \, dx = 2 \left(-2 e^{-0.5 x} + C\right) = -4 e^{-0.5 x} + 2C
\]
5. **Simplifying the constant**:
We can denote \( 2C \) as a new constant \( C' \), so we can write:
\[
-4 e^{-0.5 x} + C'
\]
Thus, the final answer is:
\[
\int 2 e^{-0.5 x} \, dx = -4 e^{-0.5 x} + C
\]
So, the exact answer is:
\[
-4 e^{-0.5 x} + C
\]
Quick Answer
\[
-4 e^{-0.5 x} + C
\]
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