Integration Of X E X-where Product Rule Thinking Returns
Integration of x e^x-where product rule thinking returns
The integral of x e^x is $$e^x(x-1)+C$$, and the cleanest method is integration by parts, which treats the problem as the product rule run in reverse. For teachers and school leaders, this is a useful example because it shows how one structured technique can turn a product into a manageable routine, rather than a memorization exercise.
Core idea
Integration by parts is built from the product rule identity $$d(uv)/dx = u(dv/dx) + v(du/dx)$$, rearranged so that one difficult integral becomes another, often simpler one. In the case of exponential growth, choosing $$u=x$$ and $$dv=e^x\,dx$$ makes the derivative smaller and the antiderivative immediate, which is exactly why this example is a standard classroom model.
- Choose $$u=x$$, so $$du=dx$$.
- Choose $$dv=e^x\,dx$$, so $$v=e^x$$.
- Apply $$\int u\,dv=uv-\int v\,du$$.
- Simplify to $$xe^x-\int e^x\,dx$$.
- Finish with $$xe^x-e^x+C$$, or equivalently $$e^x(x-1)+C$$.
Worked solution
Start with $$\int x e^x\,dx$$ and split the integrand into $$u=x$$ and $$dv=e^x\,dx$$, a choice consistent with common integration-by-parts guidance for polynomial-times-exponential forms. Then compute $$du=dx$$ and $$v=e^x$$, substitute into the formula, and obtain $$\int x e^x\,dx = xe^x - \int e^x\,dx$$, which becomes $$xe^x-e^x+C$$ after integration.
"Integration by parts is the product rule in reverse."
Step sequence
- Identify the product structure in $$x e^x$$.
- Select the algebraic factor as $$u$$.
- Integrate the remaining factor as $$dv$$.
- Substitute into $$\int u\,dv=uv-\int v\,du$$.
- Reduce the new integral and add the constant $$C$$.
Reference table
| Component | Value | Role |
|---|---|---|
| $$u$$ | $$x$$ | Differentiate this term to simplify the expression |
| $$dv$$ | $$e^x\,dx$$ | Integrate this term directly |
| $$du$$ | $$dx$$ | Derivative of $$u$$ |
| $$v$$ | $$e^x$$ | Antiderivative of $$dv$$ |
| Result | $$e^x(x-1)+C$$ | Final antiderivative |
Teaching value
This example is especially effective in secondary and pre-university math because it reinforces the link between differentiation and integration, a connection emphasized in standard calculus explanations of integration by parts. In a Marist classroom, the problem also supports disciplined reasoning: students learn to choose methods intentionally, verify each step, and present work clearly, which are habits that strengthen both mathematical competence and academic confidence.
For school systems focused on measurable learning, this type of problem is a high-yield instructional item because it tests method selection, algebraic accuracy, and symbolic fluency in one short exercise. A practical classroom sequence is to model the choice of $$u$$ and $$dv$$, let students compute the derivative and antiderivative, and then require a final check by differentiating the answer back to $$x e^x$$.
Common errors
The most common mistake is choosing $$u=e^x$$ and $$dv=x\,dx$$, which makes the new integral more awkward instead of simpler and works against the main purpose of the technique. Another frequent error is forgetting the constant $$C$$, or stopping after $$xe^x-e^x$$ without factoring to the cleaner form $$e^x(x-1)+C$$.
Implementation note
For curriculum planning, this integral is ideal for a short lesson, a guided practice set, and a retrieval quiz because it is short, elegant, and strongly tied to a transferable method. In practice, students who master this pattern are better prepared for more advanced integration tasks involving polynomial-exponential, polynomial-trigonometric, and logarithmic products.
Everything you need to know about Integration Of X E X Where Product Rule Thinking Returns
Why is the answer not just $$xe^x$$?
Because differentiating $$xe^x$$ gives $$e^x+xe^x$$, not $$xe^x$$, so the missing $$-e^x$$ term is necessary to make the derivative match the original integrand.
When should integration by parts be used?
Use it when the integrand is a product and one factor becomes simpler after differentiation while the other integrates easily, which is exactly the pattern in $$x e^x$$.
What is the final antiderivative?
The final antiderivative is $$\int x e^x\,dx = e^x(x-1)+C$$, and both this form and $$xe^x-e^x+C$$ are equivalent.
