Q. \[ \int 3e^x\,dx \]

Q: \int 3e^x\,dx ?

Use the constant multiple rule: \int 3e^x\,dx = 3\int e^x\,dx = 3e^x + C.

Q: How to integrate \int ke^x\,dx for a constant k ?

\int ke^x\,dx = k\int e^x\,dx = ke^x + C.

Q: What is \frac{d}{dx}(e^x) and how does it help?

\frac{d}{dx}(e^x)=e^x. Since derivatives match the integrand, reversing gives \int e^x\,dx=e^x + C.

Q: What is \int 3e^{2x}\,dx ?

Use substitution. \int 3e^{2x}dx = 3\cdot \frac{1}{2}e^{2x}+C=\frac{3}{2}e^{2x}+C.

Q: How to integrate \int 3e^{x+1}\,dx ?

e^{x+1}=e\cdot e^x. So \int 3e^{x+1}dx = 3e\int e^x dx = 3e\cdot e^x + C = 3e^{x+1}+C.

Q: Why do we add +C in indefinite integrals?

Because antiderivatives differ by constants: if F'(x)=f(x), then (F(x)+C)'=f(x) as well. So the general form is F(x)+C.

Answer

We want the integral of \(3e^x\). Since \(\int e^x\,dx = e^x + C\), multiplying by \(3\) gives:

\[
\int 3e^x\,dx = 3e^x + C
\]

Detailed Explanation

We want to compute the indefinite integral

\[
\int 3e^x \, dx
\]

Step 1: Use the constant multiple rule.

The factor \(3\) is a constant, so it can be pulled out of the integral.

\[
\int 3e^x \, dx = 3\int e^x \, dx
\]

Step 2: Use the basic antiderivative of \(e^x\).

A key fact is that the derivative of \(e^x\) is \(e^x\). That means the antiderivative of \(e^x\) is \(e^x\) (up to a constant).

\[
\int e^x \, dx = e^x + C
\]

Step 3: Multiply the result by \(3\).

Now multiply the antiderivative by \(3\).

\[
3\int e^x \, dx = 3\left(e^x + C\right)
\]

Step 4: Simplify the constant.

Distributing \(3\) gives

\[
3\left(e^x + C\right) = 3e^x + 3C
\]

Since \(3C\) is still just some constant, we rename it as \(C\).

\[
3e^x + 3C = 3e^x + C
\]

Final Answer:

\[
\int 3e^x \, dx = 3e^x + C
\]

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Calculus FAQ

\(\int 3e^x\,dx\) ?

Use the constant multiple rule: \(\int 3e^x\,dx = 3\int e^x\,dx = 3e^x + C\).

How to integrate \(\int ke^x\,dx\) for a constant \(k\) ?

\(\int ke^x\,dx = k\int e^x\,dx = ke^x + C\).

What is \(\frac{d}{dx}(e^x)\) and how does it help?

\(\frac{d}{dx}(e^x)=e^x\). Since derivatives match the integrand, reversing gives \(\int e^x\,dx=e^x + C\).

What is \(\int 3e^{2x}\,dx\) ?

Use substitution. \(\int 3e^{2x}dx = 3\cdot \frac{1}{2}e^{2x}+C=\frac{3}{2}e^{2x}+C\).

How to integrate \(\int 3e^{x+1}\,dx\) ?

\(e^{x+1}=e\cdot e^x\). So \(\int 3e^{x+1}dx = 3e\int e^x dx = 3e\cdot e^x + C = 3e^{x+1}+C\).

Why do we add \(+C\) in indefinite integrals?

Because antiderivatives differ by constants: if \(F'(x)=f(x)\), then \((F(x)+C)'=f(x)\) as well. So the general form is \(F(x)+C\).

Find the integral of 3e^x.
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