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Differentiation under Integral Sign

If ∫ x2· e-2x...

Question

If $\int x^{2} \cdot e^{- 2 x} d x = e^{- 2 x} (a x^{2} + b x + c) + d,$ where $d$ is the constant of integration, then the value of $∣ ∣ ∣ \frac{a}{b c} ∣ ∣ ∣$ is

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Solution

$\int x^{2} \cdot e^{- 2 x} d x = e^{- 2 x} (a x^{2} + b x + c) + d,$

Differentiating both the sides
$x^{2} \cdot e^{- 2 x} = e^{- 2 x} (2 a x + b - 2 (a x^{2} + b x + c)) \Rightarrow x^{2} = - 2 a x^{2} + (2 a - 2 b) x + b - 2 c \Rightarrow - 2 a = 1, 2 a - 2 b = 0, b - 2 c = 0 \Rightarrow a = b = - \frac{1}{2}, c = - \frac{1}{4} \Rightarrow ∣ ∣ ∣ \frac{a}{b c} ∣ ∣ ∣ = 4$

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Similar questions

\int x^{2} \cdot e^{- 2 x} d x = e^{- 2 x} (a x^{2} + b x + c) + d,

d

is the constant of integration, then the value of

∣ ∣ ∣ \frac{a}{b c} ∣ ∣ ∣

\int e^{3 x} sin 7 x d x = \frac{e^{3 x}}{a} [b sin c x + d cos c x] + C,

then the value of

a + b + c + d

(

C

is constant of integration

)

\int x^{2} \cdot e^{- 2 x} d x = e^{- 2 x} (a x^{2} + b x + c) + d,

I = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x, J = \int \frac{e^{- x}}{e^{- 4 x} + e^{- 2 x} + 1} d x .

Then the value for

J - I

equals to
( where

C

is integration constant)

\int x^{2} e^{- 2 x} d x = e^{- 2 x} (a x^{2} + b x + c) + d

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