Q. \[ \int \frac{1}{e^{x}} \, dx \]

Q: Find \displaystyle \int \frac{1}{e^x}\,dx .

\frac{1}{e^x}=e^{-x}. Then \displaystyle \int e^{-x}\,dx = -e^{-x}+C, since \frac{d}{dx}\left(-e^{-x}\right)=e^{-x}.

Q: How do you rewrite \frac{1}{e^x} as an exponential?

Use \frac{1}{e^x}=e^{-x}. More generally, \frac{a}{b}=a\,b^{-1}, and e^x=e^{x} so the reciprocal flips the exponent.

Q: What is \displaystyle \int e^{-x}\,dx ?

\displaystyle \int e^{-x}\,dx = -e^{-x}+C. The derivative of -x is -1 , which introduces the negative sign.

Q: Verify the result by differentiating -e^{-x}+C .

Differentiate: \frac{d}{dx}\left(-e^{-x}\right)= -e^{-x}\cdot(-1)=e^{-x}=\frac{1}{e^x}. The constant C differentiates to 0.

Q: What substitution method works for \displaystyle \int e^{-x}\,dx ?

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

Q: Is there a common alternative form of the antiderivative?

Yes. Since e^{-x}=\frac{1}{e^x}, the answer can be written as -\frac{1}{e^x}+C.

Answer

We want

\[
\int \frac{1}{e^x}\,dx
\]

Rewrite \( \frac{1}{e^x} = e^{-x} \).

\[
\int e^{-x}\,dx
\]

Use the rule \( \int e^{ax}\,dx = \frac{1}{a}e^{ax}+C \) with \(a=-1\):

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

Final answer:

\[
-e^{-x}+C
\]

Equivalent form:

\[
-\frac{1}{e^x}+C
\]

Detailed Explanation

We want to find the integral

\[
\int \frac{1}{e^x}\,dx.
\]

Step 1: Rewrite the integrand in a simpler exponential form.

Because \(e^x\) is in the denominator, we can use the rule

\[
\frac{1}{e^x} = e^{-x}.
\]

So the integral becomes

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

Step 2: Use the basic exponential-integral pattern.

Recall the rule:

\[
\int e^{ax}\,dx = \frac{1}{a}e^{ax} + C \quad \text{for } a \ne 0.
\]

Here, \(a = -1\), because \(e^{-x} = e^{(-1)x}\).

Step 3: Apply the rule with \(a=-1\).

Substitute \(a=-1\) into the formula:

\[
\int e^{-x}\,dx = \frac{1}{-1}e^{-x} + C.
\]

Step 4: Simplify the coefficient.

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

Step 5: (Optional) Rewrite in terms of \(e^x\) again.

Since \(e^{-x} = \frac{1}{e^x}\), we have

\[
-e^{-x} + C = -\frac{1}{e^x} + C.
\]

Final answer:

\[
\int \frac{1}{e^x}\,dx = -\frac{1}{e^x} + C.
\]

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Calculus FAQ

Find \( \displaystyle \int \frac{1}{e^x}\,dx \).

\( \frac{1}{e^x}=e^{-x}\). Then \( \displaystyle \int e^{-x}\,dx = -e^{-x}+C\), since \( \frac{d}{dx}\left(-e^{-x}\right)=e^{-x}\).

How do you rewrite \( \frac{1}{e^x} \) as an exponential?

Use \( \frac{1}{e^x}=e^{-x}\). More generally, \( \frac{a}{b}=a\,b^{-1}\), and \( e^x=e^{x}\) so the reciprocal flips the exponent.

What is \( \displaystyle \int e^{-x}\,dx \)?

\( \displaystyle \int e^{-x}\,dx = -e^{-x}+C\). The derivative of \( -x \) is \( -1 \), which introduces the negative sign.

Verify the result by differentiating \( -e^{-x}+C \).

Differentiate: \( \frac{d}{dx}\left(-e^{-x}\right)= -e^{-x}\cdot(-1)=e^{-x}=\frac{1}{e^x}\). The constant \(C\) differentiates to \(0\).

What substitution method works for \( \displaystyle \int e^{-x}\,dx \)?

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

Is there a common alternative form of the antiderivative?

Yes. Since \(e^{-x}=\frac{1}{e^x}\), the answer can be written as \( -\frac{1}{e^x}+C\).

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