Q. \[ \int e^{x/2}\,dx \]

Q: What is \\\int e^{x/2}\\,dx\?

Let \u=x/2\, so \du=dx/2\. Then \\\int e^{x/2}\\,dx=2\\int e^{u}\\,du=2e^{u}+C=2e^{x/2}+C\.

Q: How do you integrate \e^{(ax+b)}\?

Use \u=ax+b\, so \du=a\\,dx\. Then \\\int e^{ax+b}\\,dx=\\frac{1}{a}e^{ax+b}+C\ (for \a\\neq 0\).

Q: Differentiate \2e^{x/2}+C\ to check correctness.

\\\frac{d}{dx}(2e^{x/2})=2\\cdot e^{x/2}\\cdot\\frac{1}{2}=e^{x/2}\. The constant \C\ differentiates to \0\.

Q: What is \\\int_0^2 e^{x/2}\\,dx\?

Use \2e^{x/2}\\big|_0^2=2e^{1}-2e^{0}=2e-2\.

Q: What is the definite integral \\\int e^{x/2}\\,dx\ in terms of an antiderivative?

An antiderivative is \F(x)=2e^{x/2}\. For bounds \[a,b]\, \\\int_a^b e^{x/2}\\,dx=\\left[2e^{x/2}\\right]_a^b=2e^{b/2}-2e^{a/2}\.

Q: Can you integrate \\\int e^{x/2}\\,dx\ using substitution \t=x/2\?

Yes. With \t=x/2\, \dx=2dt\. Then \\\int e^{x/2}\\,dx=\\int e^{t}(2dt)=2e^{t}+C=2e^{x/2}+C\.

Answer

We want

\[
\int e^{x/2}\,dx.
\]

Let \(u=\frac{x}{2}\). Then \(du=\frac{1}{2}dx\), so \(dx=2\,du\).

\[
\int e^{x/2}\,dx=\int e^{u}\cdot 2\,du=2\int e^{u}\,du=2e^{u}+C.
\]

Substitute back \(u=\frac{x}{2}\):

\[
\int e^{x/2}\,dx=2e^{x/2}+C.
\]

Detailed Explanation

We want to compute the integral

\[
\int e^{x/2}\, dx
\]

Step 1: Identify a substitution.

The exponent is \(x/2\), so it is natural to let

\[
u=\frac{x}{2}
\]

Step 2: Compute the differential.

Differentiate both sides with respect to \(x\):

\[
\frac{du}{dx}=\frac{1}{2}
\]

\[
du=\frac{1}{2}dx
\]

Now solve for \(dx\):

\[
dx=2\,du
\]

Step 3: Rewrite the integral in terms of \(u\).

Since \(e^{x/2}=e^u\) and \(dx=2\,du\), substitute into the integral:

\[
\int e^{x/2}\, dx=\int e^{u}\,(2\,du)
\]

Factor out the constant \(2\):

\[
\int e^{x/2}\, dx=2\int e^{u}\, du
\]

Step 4: Integrate.

We use the rule

\[
\int e^{u}\, du = e^{u}+C
\]

\[
2\int e^{u}\, du = 2\left(e^{u}+C\right)=2e^{u}+C
\]

Step 5: Substitute back to \(x\).

Recall \(u=\frac{x}{2}\). Therefore

\[
2e^{u}+C=2e^{x/2}+C
\]

Final Answer:

\[
\int e^{x/2}\, dx = 2e^{x/2}+C
\]

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Calculus FAQ

What is \\(\\int e^{x/2}\\,dx\\)?

Let \\(u=x/2\\), so \\(du=dx/2\\). Then \\(\\int e^{x/2}\\,dx=2\\int e^{u}\\,du=2e^{u}+C=2e^{x/2}+C\\).

How do you integrate \\(e^{(ax+b)}\\)?

Use \\(u=ax+b\\), so \\(du=a\\,dx\\). Then \\(\\int e^{ax+b}\\,dx=\\frac{1}{a}e^{ax+b}+C\\) (for \\(a\\neq 0\\)).

Differentiate \\(2e^{x/2}+C\\) to check correctness.

\\(\\frac{d}{dx}(2e^{x/2})=2\\cdot e^{x/2}\\cdot\\frac{1}{2}=e^{x/2}\\). The constant \\(C\\) differentiates to \\(0\\).

What is \\(\\int_0^2 e^{x/2}\\,dx\\)?

Use \\(2e^{x/2}\\big|_0^2=2e^{1}-2e^{0}=2e-2\\).

What is the definite integral \\(\\int e^{x/2}\\,dx\\) in terms of an antiderivative?

An antiderivative is \\(F(x)=2e^{x/2}\\). For bounds \\([a,b]\\), \\(\\int_a^b e^{x/2}\\,dx=\\left[2e^{x/2}\\right]_a^b=2e^{b/2}-2e^{a/2}\\).

Can you integrate \\(\\int e^{x/2}\\,dx\\) using substitution \\(t=x/2\\)?

Yes. With \\(t=x/2\\), \\(dx=2dt\\). Then \\(\\int e^{x/2}\\,dx=\\int e^{t}(2dt)=2e^{t}+C=2e^{x/2}+C\\).

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