Be-Math-If-1-a-x-1-x-2-4-10-81-then-the-value-of-a-2-is-

Question Number 108761 by bemath last updated on 19/Aug/20

\frac{⋮ \frac{B e}{M a t h} ⋮}{★}

I f \int_{- 1}^{a} \frac{x + 1}{{(x + 2)}^{4}} = \frac{10}{81}, t h e n t h e v a l u e o f

a - 2 i s___

Answered by ajfour last updated on 19/Aug/20

l e t I = \int_{- 1}^{a} \frac{x + 1}{{(x + 2)}^{4}} d x

n o w s a y x + 2 = t

\Rightarrow I = \int_{1}^{a + 2} (\frac{1}{t^{3}} - \frac{1}{t^{4}}) d t = \frac{10}{81}

\Rightarrow (\frac{1}{3 t^{3}} - \frac{1}{2 t^{2}}) ∣_{1}^{a + 2} = \frac{10}{81}

\Rightarrow \frac{1}{3 {(a + 2)}^{3}} - \frac{1}{2 {(a + 2)}^{2}} = \frac{10}{81} + \frac{1}{6}

s a y \frac{1}{a + 2} = b

\Rightarrow b^{3} - \frac{3}{2} b^{2} - \frac{10}{27} - \frac{1}{2} = 0

\Rightarrow b^{3} - \frac{3}{2} b^{2} - \frac{47}{54} = 0

o r {(a + 2)}^{3} + \frac{3}{2} \times \frac{54}{47} (a + 2) - \frac{54}{47} = 0

s a y a + 2 = 3 t

\Rightarrow 47 t^{3} + 9 t - 2 = 0

\dots .

Answered by bobhans last updated on 19/Aug/20

Commented by bemath last updated on 19/Aug/20

c o o l l

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