find ∫_0 ^6 (x^2 −x+1)e^([−2x]) dx .


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Question Number 37636 by math khazana by abdo last updated on 16/Jun/18

$f i n d \int_{0}^{6} (x^{2} - x + 1) e^{[- 2 x]} d x .$
Commented by prof Abdo imad last updated on 16/Jun/18
$let A = ∫_0 ^6 (x^2 −x+1)e^([−2x]) dx A=_(2x=t) ∫_0 ^(12) ( (t^2 /4) −(t/2) +1)e^([−t]) (dt/2) = (1/8) ∫_0 ^(12) ( t^2 −2t +4) e^([−t]) dt 8A= Σ_(k=0) ^(11) ∫_k ^(k+1) (t^2 −2t +4)e^(−k) dt =Σ_(k=0) ^(11) e^(−k) ∫_k ^(k+1) (t^2 −2t +4)dt=Σ_(k=0) ^(11) e^(−k) T_k T_k = ∫_k ^(k+1) (t^2 −2t +4)dt=[(t^3 /3) −t^2 +4t]_k ^(k+1) =(((k+1)^3 )/3) −(k+1)^2 +4(k+1)−(k^3 /3) +k^2 −4k =((k^3 +3k^2 +3k +1 −k^3 )/3) −k^2 −2k −1 +k^2 +4 =k^2 +k +(1/3) −2k +3 =k^2 −k +((10)/3) ⇒8A= Σ_(k=0) ^(11) e^(−k) {k^2 −k +((10)/3)} A =(1/8)Σ_(k=0) ^(11) ( k^2 −k +((10)/3)) e^(−k) .$
$l e t A = \int_{0}^{6} (x^{2} - x + 1) e^{[- 2 x]} d x$ $A =_{2 x = t} \int_{0}^{12} (\frac{t^{2}}{4} - \frac{t}{2} + 1) e^{[- t]} \frac{d t}{2}$ $= \frac{1}{8} \int_{0}^{12} (t^{2} - 2 t + 4) e^{[- t]} d t$ $8 A = \sum_{k = 0}^{11} \int_{k}^{k + 1} (t^{2} - 2 t + 4) e^{- k} d t$ $= \sum_{k = 0}^{11} e^{- k} \int_{k}^{k + 1} (t^{2} - 2 t + 4) d t = \sum_{k = 0}^{11} e^{- k} T_{k}$ $T_{k} = \int_{k}^{k + 1} (t^{2} - 2 t + 4) d t = {[\frac{t^{3}}{3} - t^{2} + 4 t]}_{k}^{k + 1}$ $= \frac{{(k + 1)}^{3}}{3} - {(k + 1)}^{2} + 4 (k + 1) - \frac{k^{3}}{3} + k^{2} - 4 k$ $= \frac{k^{3} + 3 k^{2} + 3 k + 1 - k^{3}}{3} - k^{2} - 2 k - 1 + k^{2} + 4$ $= k^{2} + k + \frac{1}{3} - 2 k + 3$ $= k^{2} - k + \frac{10}{3} \Rightarrow 8 A = \sum_{k = 0}^{11} e^{- k} {k^{2} - k + \frac{10}{3}}$ $A = \frac{1}{8} \sum_{k = 0}^{11} (k^{2} - k + \frac{10}{3}) e^{- k} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18

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