Question-155382

Question Number 155382 by MathsFan last updated on 29/Sep/21

Answered by puissant last updated on 29/Sep/21

Q = \int_{0}^{10} \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots}}}} d x

u = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots .}}}} \to u^{2} = x + \sqrt{x + \sqrt{x + \sqrt{x + \dots .}}}

\Rightarrow u^{2} - u = x \Rightarrow u - u + \frac{1}{4} = x + \frac{1}{4}

\Rightarrow {(u - \frac{1}{2})}^{2} = x + \frac{1}{4} \Rightarrow u = \frac{1}{2} + \sqrt{\frac{4 x + 1}{4}}

Q = \int_{0}^{10} (\frac{1}{2} + \frac{1}{2} \sqrt{4 x + 1}) d x = \frac{1}{2} \int_{0}^{10} (1 + \sqrt{4 x + 1}) d x

= \frac{1}{2} {[x + \frac{2}{3} \times \frac{1}{4} \sqrt{{(4 x + 1)}^{3}}]}_{0}^{10} = \frac{1}{2} {[x + \frac{1}{6} \sqrt{{(4 x + 1)}^{3}}]}_{0}^{10}

= \frac{1}{2} [(10 + \frac{\sqrt{68921}}{6}) - (0 + \frac{1}{6})]

∴ ∵ Q = \frac{1}{12} (59 + 2 \sqrt{68921}) . .

\dots \dots \dots \dots . . L e p u i s s a n t \dots \dots \dots \dots \dots

Commented by SANOGO last updated on 30/Sep/21

l e d u r g a r

Commented by MathsFan last updated on 30/Sep/21

c o r r e c t s i r

Commented by peter frank last updated on 01/Oct/21

good