2. no calculator\n$\\frac{1}{2}\\int e^{\\frac{t}{2}}dt=$\n(a) $e^{-t}+c$ (b) $e^{-\\frac{t}{2}}+c$ (c)…

2. no calculator\n$\\frac{1}{2}\\int e^{\\frac{t}{2}}dt=$\n(a) $e^{-t}+c$ (b) $e^{-\\frac{t}{2}}+c$ (c) $e^{\\frac{t}{2}}+c$ (d) $2e^{\\frac{t}{2}}+c$ (e) $e^{t}+c$
Answer
Explanation:
Step1: Use substitution method
Let (u = \frac{t}{2}), then (du=\frac{1}{2}dt).
Step2: Substitute into the integral
The original integral (\frac{1}{2}\int e^{\frac{t}{2}}dt) becomes (\int e^{u}du).
Step3: Integrate (e^{u})
Since (\int e^{u}du=e^{u}+C).
Step4: Substitute back (u = \frac{t}{2})
We get (e^{\frac{t}{2}}+C).
Answer:
C. (e^{\frac{t}{2}}+C)