Student solution · Integration by parts
Evaluate ∫ x²eˣ dx.
I = ∫ x²eˣ dx
u = x², dv = eˣ dx
du = 2x dx, v = eˣ
I = x²eˣ − ∫ 2xeˣ dx
∫ 2xeˣ dx = 2xeˣ − ∫ 2eˣ dx
= 2xeˣ − 2eˣ
I = x²eˣ − (2xeˣ − 2eˣ)
= x²eˣ − 2xeˣ − 2eˣ + C
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Rubric breakdownAnalyzing student work…
✓
Choose a valid setup\(u=x^2,\;dv=e^x\,dx\)
1 / 1✓
First integration by parts\(I=x^2e^x-\int 2xe^x\,dx\)
2 / 2✓
Second integration by parts\(\int 2xe^x\,dx=2xe^x-\int 2e^x\,dx\)
2 / 2✓
Evaluate intermediate result\(\int 2xe^x\,dx=2xe^x-2e^x\)
2 / 2✓
Substitute back\(I=x^2e^x-(2xe^x-2e^x)\)
1 / 1Final answerExpected: \(x^2e^x-2xe^x+2e^x+C\)
0 / 2Method substantially correct
8/10Teacher can review and edit the score.
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