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प्रश्न
Evaluate the following integral:
`int_(-1)^1 x^2 "e"^(-2x) "d"x`
उत्तर
`int_(-1)^1 x^2 "e"^(-2x) "d"x`
We use integration by parts
Let u = x2
Then du = 2x dx
Let dv = `"e"^(-2x) "d"x`
v = `"e"^(-2x)/(-2)`
= `[(x^2"e"^(-2x))/(-2)]_(-1)^1 - int_(-1)^1 (2x) "e"^(-2x)/(-2) "d"x`
= `"e"^(-2x)/(-2) - "e"^2/(-2) + int_(-1)^1 x"e"^(-2x) "d"x`
= `("e"^2 - "e"^(-2))/2 + [(x"e"^(-2x))/(-2)]_(-1)^1 - int_(-1)^1 "e"^(-2x)/(-2) "d"x`
We have used itegration by prts again
= `("e"^2 - "e"^(-2))/2 - "e"^(-2)/2 - "e"^2/2 + 1/2["e"^(-2x)/(-2)]_(-1)^1`
= `-"e"^(-2) + 1/2 ["e"^(-2)/(-2) + "e"^2/2]`
= `(-1)/"e"^2 - 1/(4"e"^2) + 1/4 "e"^2`
= `"e"^2/4 - 5/(4"e"^2)`
= `1/4 [("e"^4 - 5)/"e"^2]`
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