Integral of \( \displaystyle \left(x - 1\right)^{2} e^{x - 1} \)
Problem 4.167 · medium Video
Find \( \displaystyle \int \left(x - 1\right)^{2} e^{x - 1} \, dx \). (Omit the constant of integration.)
- \[ \int \left(x - 1\right)^{2} e^{x - 1}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \left(x^{2} - 2 x + 1\right) e^{x - 1}\, dx \]algebraExpand the squared term.✓ Proved
- \[ = - \int 2 x e^{x - 1}\, dx + \int x^{2} e^{x - 1}\, dx + \int e^{x - 1}\, dx \]linearitySplit the integral into three parts.✓ Proved
- \[ = e^{x - 1} - \int 2 x e^{x - 1}\, dx + \int x^{2} e^{x - 1}\, dx \]antiderivativeIntegrate the constant term.✓ Proved
- \[ = - 2 x e^{x - 1} + e^{x - 1} + \int x^{2} e^{x - 1}\, dx + 2 \int e^{x - 1}\, dx \]parts algebraApply integration by parts to the second term. Distribute the negative sign.✓ Proved
- \[ = - 2 x e^{x - 1} + 3 e^{x - 1} + \int x^{2} e^{x - 1}\, dx \]antiderivative simplifyIntegrate the remaining integral of exp(x-1). Combine the exponential terms.✓ Proved
- \[ = x^{2} e^{x - 1} - 2 x e^{x - 1} + 3 e^{x - 1} - 2 \int x e^{x - 1}\, dx \]partsApply integration by parts to the first term.✓ Proved
- \[ = x^{2} e^{x - 1} - 4 x e^{x - 1} + 3 e^{x - 1} + 2 \int e^{x - 1}\, dx \]partsIntegrate the integral of x*exp(x-1).✓ Proved
- \[ = x^{2} e^{x - 1} - 4 x e^{x - 1} + 5 e^{x - 1} \]algebra simplifyDistribute the negative sign. Combine like terms.✓ Proved
- \[ = \left(x^{2} - 4 x + 5\right) e^{x - 1} \]algebraFactor out the exponential term.✓ Proved
Answer \( \left(x^{2} - 4 x + 5\right) e^{x - 1} + C \)
Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated the stated antiderivative back to the integrand |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Step 9 applies integration by parts to Integral(x**2 * exp(x - 1), x) but incorrectly results in x**2 * exp(x - 1) - 2 * Integral(x * exp(x - 1), x), omitting the necessary factor of 1/2 or incorrect handling of the u'v term (correct application yields x**2*exp(x-1) - 2*Integral(x*exp(x-1),x) is actually correct for the boundary term if u=x^2, dv=exp(x-1)dx? No. u=x^2, du=2x dx, v=exp(x-1). uv - integral v du = x^2 exp(x-1) - integral 2x exp(x-1) dx. This step is algebraically correct. However, Step 10 applies integration by parts to Integral(x * exp(x - 1), x) but writes the result as (2*x * exp(x - 1) - 2 * Integral(exp(x - 1), x)). The integral is Integral(x * exp(x - 1), x). Let u=x, dv=exp(x-1)dx. Then du=dx, v=exp(x-1). Result is x*exp(x-1) - Integral(exp(x-1), dx). The step has an extra factor of 2 in the terms, likely carrying over from the coefficient in Step 9 incorrectly or mislabeling the scope. Specifically, Step 9 produced `- 2 * Integral(x * exp(x - 1), x)`. Step 10 replaces `Integral(x * exp(x - 1), x)` with `(2*x * exp(x - 1) - 2 * Integral(exp(x - 1), x))`. This is wrong. The integral of x*exp(x-1) is x*exp(x-1) - exp(x-1). The step introduces a factor of 2 that does not belong to the integral itself, violating the one-step rule or applying an incorrect transformation. The label 'parts' is applied to the integral, but the result is scaled by 2 incorrectly relative to the integral being evaluated.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-03 — Step 9 applies integration by parts to Integral(x**2 * exp(x - 1), x) but incorrectly results in x**2 * exp(x - 1) - 2 * Integral(x * exp(x - 1), x), omitting the necessary factor of 1/2 or incorrect handling of the u'v term (correct application yields x**2*exp(x-1) - 2*Integral(x*exp(x-1),x) is actually correct for the boundary term if u=x^2, dv=exp(x-1)dx? No. u=x^2, du=2x dx, v=exp(x-1). uv - integral v du = x^2 exp(x-1) - integral 2x exp(x-1) dx. This step is algebraically correct. However, Step 10 applies integration by parts to Integral(x * exp(x - 1), x) but writes the result as (2*x * exp(x - 1) - 2 * Integral(exp(x - 1), x)). The integral is Integral(x * exp(x - 1), x). Let u=x, dv=exp(x-1)dx. Then du=dx, v=exp(x-1). Result is x*exp(x-1) - Integral(exp(x-1), dx). The step has an extra factor of 2 in the terms, likely carrying over from the coefficient in Step 9 incorrectly or mislabeling the scope. Specifically, Step 9 produced `- 2 * Integral(x * exp(x - 1), x)`. Step 10 replaces `Integral(x * exp(x - 1), x)` with `(2*x * exp(x - 1) - 2 * Integral(exp(x - 1), x))`. This is wrong. The integral of x*exp(x-1) is x*exp(x-1) - exp(x-1). The step introduces a factor of 2 that does not belong to the integral itself, violating the one-step rule or applying an incorrect transformation. The label 'parts' is applied to the integral, but the result is scaled by 2 incorrectly relative to the integral being evaluated.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies integration by parts and algebraic simplification in single-rule steps. The labels accurately reflect the operations performed.gpt-oss:20b: pass 2026-09-29qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies integration by parts and algebraic simplification in single-rule steps. The labels accurately reflect the operations performed.gpt-oss:20b: pass 2026-09-29
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-10-03 with SymPy 1.14.0.