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

Q: How do I compute \int x e^{x}\,dx using integration by parts?

Let u=x, dv=e^x dx. Then du=dx, v=e^x. So \int x e^x dx = x e^x-\int e^x dx = x e^x-e^x + C = e^x(x-1)+C.

Q: What should I choose for u and dv in \int x e^x dx ?

Choose u=x (polynomial part) and dv=e^x dx because e^x stays simple when integrated. This makes \int e^x dx easy and leads quickly to e^x(x-1)+C.

Q: Can I verify the result by differentiating e^x(x-1)?

Differentiate: \frac{d}{dx}[e^x(x-1)] = e^x(x-1)+e^x = e^x x. So the derivative is x e^x, confirming \int x e^x dx = e^x(x-1)+C.

Q: What is \int (x-1)e^{x}\,dx ?

Since (x-1)e^x matches the factor in the antiderivative, use the result: \int (x-1)e^x dx = e^x(x-1-0?) + C. More directly, \int (x-1)e^x dx = e^x(x-2)+C by applying the same pattern.

Q: What is the definite integral \int_{0}^{1} x e^{x}\,dx ?

Use e^x(x-1). Evaluate: [e^x(x-1)]_{0}^{1} = e(1-1)-1(0-1)=0-(-1)=1. So the value is 1.

Q: How do I handle \int x e^{x}\,dx using the product rule in reverse?

Notice \frac{d}{dx}[x e^x] = e^x + x e^x. Rearranging gives x e^x = \frac{d}{dx}[x e^x]-e^x. Integrate: \int x e^x dx = x e^x - e^x + C.

Q: What about \int x e^{x} dx with a constant factor, like \int 3x e^{x} dx ?

Use linearity. From \int x e^x dx = e^x(x-1)+C, multiply by 3: \int 3x e^x dx = 3e^x(x-1)+C.

Answer

Use integration by parts. Let \(u=x\) so \(du=dx\), and let \(dv=e^x\,dx\) so \(v=e^x\).

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

Final result:

\[
\int x e^x\,dx = e^x(x-1) + C
\]

Detailed Explanation

We want to compute the indefinite integral

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

Because this is a product of a polynomial \(x\) and an exponential \(e^x\), the standard method is integration by parts.

Step 1: Choose \(u\) and \(dv\)

Integration by parts uses the formula

\[
\int u \, dv = u v – \int v \, du
\]

Choose:

\(u = x\)
\(dv = e^x \, dx\)

Now compute the derivatives and integrals needed for the formula.

Step 2: Compute \(du\) and \(v\)

Differentiate \(u\):

\[
u = x \quad \Rightarrow \quad du = dx
\]

Integrate \(dv\) to get \(v\):

\[
dv = e^x \, dx \quad \Rightarrow \quad v = \int e^x \, dx = e^x
\]

Step 3: Apply the integration by parts formula

Substitute into

\[
\int u \, dv = u v – \int v \, du
\]

This gives:

\[
\int x e^x \, dx = x \cdot e^x – \int e^x \cdot dx
\]

Step 4: Integrate the remaining integral

Now evaluate:

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

So the expression becomes:

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

Step 5: Simplify (factor if possible)

Factor out \(e^x\) from the terms \(x e^x – e^x\):

\[
x e^x – e^x = e^x(x – 1)
\]

Therefore, the final answer is:

\[
\int x e^x \, dx = e^x(x – 1) + C
\]

See full solution

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Calculus FAQ

How do I compute \( \int x e^{x}\,dx \) using integration by parts?

Let \(u=x\), \(dv=e^x dx\). Then \(du=dx\), \(v=e^x\). So \(\int x e^x dx = x e^x-\int e^x dx = x e^x-e^x + C = e^x(x-1)+C\).

What should I choose for \(u\) and \(dv\) in \( \int x e^x dx \)?

Choose \(u=x\) (polynomial part) and \(dv=e^x dx\) because \(e^x\) stays simple when integrated. This makes \(\int e^x dx\) easy and leads quickly to \(e^x(x-1)+C\).

Can I verify the result by differentiating \(e^x(x-1)\)?

Differentiate: \(\frac{d}{dx}[e^x(x-1)] = e^x(x-1)+e^x = e^x x\). So the derivative is \(x e^x\), confirming \(\int x e^x dx = e^x(x-1)+C\).

What is \( \int (x-1)e^{x}\,dx \)?

Since \((x-1)e^x\) matches the factor in the antiderivative, use the result: \(\int (x-1)e^x dx = e^x(x-1-0?) + C\). More directly, \(\int (x-1)e^x dx = e^x(x-2)+C\) by applying the same pattern.

What is the definite integral \( \int_{0}^{1} x e^{x}\,dx \)?

Use \(e^x(x-1)\). Evaluate: \([e^x(x-1)]_{0}^{1} = e(1-1)-1(0-1)=0-(-1)=1\). So the value is \(1\).

How do I handle \( \int x e^{x}\,dx \) using the product rule in reverse?

Notice \( \frac{d}{dx}[x e^x] = e^x + x e^x\). Rearranging gives \(x e^x = \frac{d}{dx}[x e^x]-e^x\). Integrate: \(\int x e^x dx = x e^x - e^x + C\).

What about \( \int x e^{x} dx \) with a constant factor, like \( \int 3x e^{x} dx \)?

Use linearity. From \(\int x e^x dx = e^x(x-1)+C\), multiply by \(3\): \(\int 3x e^x dx = 3e^x(x-1)+C\).

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