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1. (a) Find \( \int x^2 e^x \, dx \) - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 9
Question 3
1. (a) Find \( \int x^2 e^x \, dx \). (b) Hence find the exact value of \( \int_0^1 x^2 e^x \, dx \).
Worked Solution & Example Answer:1. (a) Find \( \int x^2 e^x \, dx \) - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 9
Step 1
Find \( \int x^2 e^x \, dx \)
Answer
To find ( \int x^2 e^x , dx ), we will use integration by parts. Let:
- ( u = x^2 ) implies ( du = 2x , dx )
- ( dv = e^x , dx ) implies ( v = e^x )
Using the integration by parts formula ( \int u , dv = uv - \int v , du ), we have:
[ \int x^2 e^x , dx = x^2 e^x - \int e^x (2x) , dx. ]
Now we need to compute ( \int 2x e^x , dx ) using integration by parts again:
Let:
- ( u = 2x ) implies ( du = 2 , dx )
- ( dv = e^x , dx ) implies ( v = e^x )
Thus,
[ \int 2x e^x , dx = 2x e^x - \int 2 e^x , dx. ]
Calculating ( \int 2 e^x , dx ) yields:
[ 2e^x. ]
Now, substituting back, we find:
[ \int x^2 e^x , dx = x^2 e^x - (2x e^x - 2e^x) + C \Rightarrow x^2 e^x - 2x e^x + 2e^x + C. ]
Step 2
Hence find the exact value of \( \int_0^1 x^2 e^x \, dx \)
Answer
Now, substituting the limits of integration from 0 to 1 into ( \int x^2 e^x , dx ):
[ \int_0^1 x^2 e^x , dx = \left[ x^2 e^x - 2x e^x + 2e^x \right]_0^1. ]
Calculating at the upper limit (when ( x = 1 )):
[ 1^2 e^1 - 2(1)(e^1) + 2e^1 = e - 2e + 2e = e. ]
At the lower limit (when ( x = 0 )):
[ 0^2 e^0 - 2(0)(e^0) + 2e^0 = 0 - 0 + 2 = 2. ]
Therefore, combining both results gives:
[ \int_0^1 x^2 e^x , dx = e - 2. ]