It becomes child's play, part 2 (My eighth integral problem)

Calculus Level 5

∫ 2 3 [ x 2 + x + 1 ] − [ x 2 + x + 1 ] + [ x 2 + x + 1 ] − [ x 2 + x + 1 ] + ⋯ d x \int_{2}^{3} \sqrt{\left[x^{2} + x + 1 \right]-\sqrt{\left[x^{2} + x + 1\right]+\sqrt{\left[x^{2} + x + 1 \right]-\sqrt{\left[x^{2} + x + 1 \right]+\cdots}}}} \, dx

If the value of the integral above is V V , input 10 V 10V as the answer.


The answer is 25.

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Hobart Pao by Hobart Pao , 23, USA

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3 solutions

Rishabh Jain
Feb 8, 2016

x = x 2 = x 2 + x + 1 − ( x + 1 ) x=\sqrt{x^2}=\sqrt{x^2+x+1-\color{#EC7300}{(x+1)}} = x 2 + x + 1 − ( x + 1 ) 2 =\sqrt{x^2+x+1- \sqrt{\color{#EC7300}{(x+1)^2}}} = x 2 + x + 1 − x 2 + x + 1 + x =\sqrt{x^2+x+1- \sqrt{{x^2+x+1+\color{#D61F06}{x}}}} Writing x a s x 2 \color{#D61F06}{x} ~as~\sqrt{x^2} and following the same pattern from 1st step generates the required pattern ........
Hence,
∫ 2 3 x d x = 1 2 ( x 2 ∣ 2 3 ) \Large \int_{2}^3 xdx=\dfrac{1}{2}(x^2|_{\small{2}}^{\small{3}}) = 5 2 =\dfrac{5}{2} ∴ 10 × 5 2 = 25 \huge\therefore 10\times\dfrac{5}{2}=\boxed{\color{#007fff}{25}}

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Very nice! That's another way to deal with those nested radicals! I didn't think of it that way! Thanks!

Hobart Pao - 5 years, 4 months ago

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T h a n k s ! \Large{\color{#302B94}{\mathbb{Thanks!}}}

Rishabh Jain - 5 years, 4 months ago

Your solution depends on you knowing that the expression is equal to x x .

So here is what one can obtain without knowing it.

Let us first try to generalize a few things,

a + a − a + a − … = m a − a + a − a + … = n ⇒ { a + n = m a − m = n \sqrt{ a+\sqrt{ a- \sqrt{a+ \sqrt{a-\ldots } }}}=m \\ \sqrt{ a-\sqrt{ a+ \sqrt{a- \sqrt{a+\ldots } }}}=n \\\Rightarrow \begin{cases} \sqrt{a+n}=m \\ \sqrt{a-m}=n \end{cases}

Squaring both the equations,

a + n = m 2 a − m = n 2 m 2 − n 2 = m + n m − n = 1 a+n=m^2 \\ a-m=n^2 \\ m^2-n^2=m+n \\ m-n=1

Now placing m = a + n m=\sqrt{a+n} in and squaring,

a + n = n 2 + 2 n + 1 n 2 + n + 1 − a = 0 n = − 1 + 1 − 4 ( 1 − a ) 2 a+n=n^2+2n+1 \\ n^2+n+1-a=0 \\n=\frac{-1+\sqrt{1-4(1-a)}}{2}

Similarly,

m = 1 + 1 − 4 ( 1 − a ) 2 m= \frac{1+\sqrt{1-4(1-a)}}{2}

So, in our original question, a = x 2 + x + 1 a=x^2+x+1 and therefore,

n = − 1 + 1 − 4 ( 1 − x 2 − x − 1 ) 2 = − 1 + 4 x 2 + 4 x + 1 2 = 2 x 2 = x \begin{aligned} n & = \frac{-1+\sqrt{1-4(1-x^2-x-1)}}{2} \\ &= \frac{-1+\sqrt{4x^2+4x+1}}{2} \\ &= \frac{2x}{2}=x \end{aligned}

Akshat Sharda - 5 years, 4 months ago

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I also didn't knew that the expression would turn into x... But you have to think, analyse, manipulate and do much more else to see that there's a pattern hidden in the question turning it into something simple...indeed very simple{just x.. :) }. Anyways nice...

Rishabh Jain - 5 years, 4 months ago

Let y y be the value of the repeated radical, and z z the value of the first embedded radical, then y 2 = x 2 + x + 1 − z ; z 2 = x 2 + x + 1 + y . y^2 = x^2 + x + 1 - z;\ \ \ \ z^2 =x^2 + x + 1 + y. Subtraction shows [z^2 - y^2 = z + y\ \ \therefore\ \ (z+y)(z-y) = z+y\ \ \therefore\ \ (z+y)(z-y - 1) = 0\ \ \therefore z = − y z = -y or z − y − 1 = 0 z - y - 1 = 0 .

We can rule out the former, because z z and y y must both be positive. This leaves z = y + 1 z = y+1 so that y 2 = x 2 + x − y ∴ y 2 − x 2 = x − y ∴ ( y − x ) ( x + y + 1 ) = 0 , y^2 = x^2 + x - y\ \ \therefore\ \ y^2 - x^2 = x-y\ \ \therefore\ \ (y-x)(x+y+1) = 0, so that either y = x y = x and z = x + 1 z = x+1 , or y = − x − 1 y = -x-1 and z = − x z = -x .

The latter option works is x x is negative, which is not the case here. Thus we are left with

x 2 + x + 1 − x 2 + x + 1 + ⋯ = x 2 + x + 1 − x 2 + x + 1 + x = x 2 + x + 1 − ( x + 1 ) = x 2 = x , \sqrt{x^2 + x + 1 - \sqrt{x^2 + x + 1 + \cdots}} = \sqrt{x^2 + x + 1 - \sqrt{x^2 + x + 1 + x}} \\ = \sqrt{x^2 + x + 1 - (x+1)} = \sqrt{x^2} = x, provided that x ≥ 0 x \geq 0 .

Finish up the problem is now easy. ∫ 2 3 x d x = 1 2 x 2 ∣ 2 3 = 1 2 ( 3 2 − 2 2 ) = 2 1 2 , \int_2^3 x\:dx = \left.\tfrac12x^2\right|_2^3 = \tfrac12(3^2 - 2^2) = 2\frac12, so we report the value 25 \boxed{25} .

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Let { u = x 2 + x + 1 − x 2 + x + 1 + x 2 + x + 1 − x 2 + x + 1 + ⋯ t = x 2 + x + 1 + x 2 + x + 1 − x 2 + x + 1 + x 2 + x + 1 − ⋯ \begin{cases} u = \sqrt{x^2+x+1 - \sqrt{x^2+x+1 + \sqrt{x^2+x+1 - \sqrt{x^2+x+1 + \cdots }}}} \\ t = \sqrt{x^2+x+1 + \sqrt{x^2+x+1 - \sqrt{x^2+x+1 + \sqrt{x^2+x+1 - \cdots}}}} \end{cases}

⟹ { u 2 = x 2 + x + 1 − t . . . ( 1 ) t 2 = x 2 + x + 1 + u . . . ( 2 ) \implies \begin{cases} u^2 = x^2+x+1-t & ...(1) \\ t^2 = x^2+x+1 + u & ...(2) \end{cases} ⟹ ( 1 ) − ( 2 ) : u 2 − t 2 = − t − u ⟹ ( u − t ) ( u + t ) = − ( u + t ) ⟹ u − t = − 1 ⟹ t = u + 1 \implies (1)-(2): u^2-t^2 = - t - u \implies (u-t)(u+t) = -(u+t) \implies u-t = - 1 \implies t = u+1

⟹ ( 1 ) : u 2 = x 2 + x + 1 − ( u + 1 ) ⟹ u 2 + u = x 2 + x ⟹ u = x \implies (1): u^2 = x^2+x+1- (u+1) \implies u^2 + u = x^2 + x \implies u = x .

Therefore, ∫ 2 3 u d x = ∫ 2 3 x d x = x 2 2 ∣ 2 3 = 3 2 − 2 2 2 = 2.5 \displaystyle \int_2^3 u \ dx = \int_2^3 x \ dx = \frac {x^2}2 \bigg|_2^3 = \frac {3^2-2^2}2 = 2.5 . ⟹ 10 V = 10 × 2.5 = 25 \implies 10V = 10 \times 2.5 = \boxed{25} .

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