If - x 6 - 7 x 5 - 6 x 4 - 5 x 3 - 4 x 2 - 3 x - 1 e x d x is equal to - k - 1 - x k - e x - c where C is constant of integration then - - - is-

The correct option is B 7 ∫ (x^6+7x^5+6x^4+5x^3+4x^2+3x+1) e^xdx= ∑_k = 1^αβ x^k. e^x+c =x^6e^x ∫ 6x^5e^xd+ ∫ (7x^5+6x^4+5x^3+4x^2+3x+1) e^x dx ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If ∫ x 6 +...

Question

If $\int (x^{6} + 7 x^{5} + 6 x^{4} + 5 x^{3} + 4 x^{2} + 3 x + 1) e^{x} d x$ is equal to $\sum_{k = 1}^{\infty} β x^{k} \cdot e^{x} + c$ (where C is constant of integration) then $(α + β)$ is-

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Solution

The correct option is B $7$
$\int (x^{6} + 7 x^{5} + 6 x^{4} + 5 x^{3} + 4 x^{2} + 3 x + 1) e^{x} d x = \sum_{k = 1}^{α} β x^{k} . e^{x} + c$
$= x^{6} e^{x} - \int 6 x^{5} e^{x} d + \int (7 x^{5} + 6 x^{4} + 5 x^{3} + 4 x^{2} + 3 x + 1) e^{x} d x$
$= x^{6} e^{x} + \int (x^{5} + 6 x^{4} + 5 x^{3} + 4 x^{2} + 3 x + 1) e^{x} d x$
$= x^{6} e^{x} + x^{5} e^{x} + \int (x^{4} + 5 x^{3} + 4 x^{2} + 3 x + 1) e^{x} d x$
$= x^{6} e^{x} + x^{5} + e^{x} + x^{4} e^{x} + \int (x^{3} + 4 x^{2} + 3 x + 1) e^{x} d x$
$= x^{6} e^{x} + x^{5} e^{x} + x^{4} e^{x} + x^{3} e^{x} + \int (x^{2} + 3 x + 1) e^{x} d x$
$= x^{6} e^{x} + x^{5} e^{x} + x^{4} e^{x} + x^{3} e^{x} + x^{2} e^{x} + x e^{x} = \sum_{k = 1}^{6} . x^{k} e^{x} + c$
$α = 6 β = 1 \Rightarrow α + β = 6 = 7$ .

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If ∫(x6+7x5+6x4+5x3+4x2+3x+1)exdx is equal to ∑∞k=1βxk⋅ex+c (where C is constant of integration) then (α+β) is-

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