Student solution · Integration by parts
Evaluate ∫ x²eˣ dx.
u = x², dv = eˣ dx
du = 2x dx, v = eˣ
∫x²eˣdx = x²eˣ − ∫2xeˣdx
∫2xeˣdx = 2xeˣ − 2eˣ
∫x²eˣdx = x²eˣ − 2xeˣ − 2eˣ + C
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Calculus I · MidtermTeacher review
Evaluate ∫ x²eˣ dx using integration by parts.
∫ x²eˣ dx = x²eˣ − ∫2xeˣ dx
= x²eˣ − (2xeˣ − 2eˣ)
= x²eˣ − 2xeˣ − 2eˣ + C
= x²eˣ − (2xeˣ − 2eˣ)
= x²eˣ − 2xeˣ − 2eˣ + C
Method selected✓
First integration✓
Second integration✓
Substitution△
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Rubric breakdownAnalyzing student work…
✓
Choose a valid setupCorrect choice of u and dv
1 / 1✓
First integration by partsStructure is correct
2 / 2✓
Second integration by partsIntermediate integral is correct
2 / 2✓
Evaluate intermediate result2xeˣ − 2eˣ is correct
2 / 2△
Substitute backSign error when distributing the minus
1 / 2△
Final answerConstant-term sign is incorrect
0 / 1Method substantially correct
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Calculus IMidtermLaTeXPDF
Calculus I — Midterm Examination
1. Evaluate ∫ x²eˣ dx using integration by parts. Show your work.
2. Determine the interval of convergence of the power series …
3. Find the area enclosed by the curves …
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AssessmentCourseAverageStatus
Midterm 1Calculus I82.4Reviewed
Quiz 4Calculus I76.8Archived
FinalEngineering Math84.1Reviewed
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