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

Q: Why does \int e^{ax}\,dx = \frac{1}{a}e^{ax}+C for a\neq 0?

Differentiate \frac{1}{a}e^{ax} : \frac{d}{dx}\left(\frac{1}{a}e^{ax}\right)=e^{ax} .

Q: Find \int e^{-x}\,dx using substitution u=-x.

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

Answer

To integrate \(e^{-x}\), use the rule \(\int e^{ax}\,dx=\frac{1}{a}e^{ax}+C\) for \(a\neq 0\). Here \(a=-1\).

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

Final result: \(-e^{-x}+C\).

Detailed Explanation

We want to compute the indefinite integral

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

Step 1: Choose a substitution.

Inside the exponent we have \(-x\). A standard way to integrate exponentials is to let the exponent be the new variable.

Let

\[
u=-x
\]

Step 2: Differentiate the substitution.

If \(u=-x\), then differentiating both sides gives

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

\[
du=-dx
\]

Equivalently, multiply both sides by \(-1\) to rewrite \(dx\) in terms of \(du\):

\[
dx=-du
\]

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

Substitute \(u=-x\) and \(dx=-du\) into the integral:

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

Factor out the negative sign:

\[
\int e^{u}\,(-du) = -\int e^{u}\,du
\]

Step 4: Integrate with respect to \(u\).

We use the basic rule

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

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

Thus we get

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

Step 5: Substitute back \(u=-x\).

Replace \(u\) with \(-x\):

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

Final Answer.

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

See full solution

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Calculus FAQ

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

\( \int e^{-x}\,dx = -e^{-x}+C \).

Why does \( \int e^{ax}\,dx = \frac{1}{a}e^{ax}+C \) for \(a\neq 0\)?

Differentiate \( \frac{1}{a}e^{ax} \): \( \frac{d}{dx}\left(\frac{1}{a}e^{ax}\right)=e^{ax} \).

Compute \( \int -e^{-x}\,dx \).

\( \int -e^{-x}\,dx = e^{-x}+C \).

Evaluate \( \int_{0}^{\infty} e^{-x}\,dx \).

\( \int_{0}^{\infty} e^{-x}\,dx = \left[-e^{-x}\right]_{0}^{\infty}=0-(-1)=1 \).

Find \( \int e^{-x}\,dx \) using substitution \(u=-x\).

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

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