Question

The value of the integral $\int_0^1 \sqrt{x + \sqrt{x + \sqrt{x + ...}}} dx$ is:

• A. $\frac{1}{12}(\sqrt{3} + 1)$
• B. $\frac{5}{12}(\sqrt{3} + 1)$
• C. $\frac{5}{12}(\sqrt{5} + 1)$
• D. $\frac{1}{12}(\sqrt{5} + 1)$

Answer 1

Answer: (C)

$\frac{5}{12}(\sqrt{5} + 1)$

Answer 2

To evaluate the integral $\int_0^1 \sqrt{x + \sqrt{x + \sqrt{x + ...}}} dx$, we first need to simplify the expression inside the integral.

Let $y = \sqrt{x + \sqrt{x + \sqrt{x + ...}}}$. This is an infinite nested radical. We can express it as:

$y = \sqrt{x + y}$

Now, square both sides of the equation:

$y^2 = x + y$

Rearrange the terms to form a quadratic equation in $y$:

$y^2 - y - x = 0$

Solve for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-1$, and $c=-x$:

$y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-x)}}{2(1)}$

$y = \frac{1 \pm \sqrt{1 + 4x}}{2}$

Since $y$ represents a square root, it must be non-negative. The term $\sqrt{1+4x}$ is always positive for $x \ge 0$. If we take the minus sign, $1 - \sqrt{1+4x}$ would be negative (because $\sqrt{1+4x} > 1$ for $x > 0$). Therefore, we must choose the positive root:

$y = \frac{1 + \sqrt{1 + 4x}}{2}$

Now, substitute this expression for $y$ back into the integral:

$I = \int_0^1 \frac{1 + \sqrt{1 + 4x}}{2} dx$

We can factor out $\frac{1}{2}$ and split the integral:

$I = \frac{1}{2} \int_0^1 (1 + \sqrt{1 + 4x}) dx$

$I = \frac{1}{2} \left[ \int_0^1 1 dx + \int_0^1 (1 + 4x)^{1/2} dx \right]$

Evaluate each integral separately:

1. $\int_0^1 1 dx = [x]_0^1 = 1 - 0 = 1$

2. $\int_0^1 (1 + 4x)^{1/2} dx$

To integrate this, use a substitution. Let $u = 1 + 4x$. Then $du = 4 dx$, which means $dx = \frac{1}{4} du$.

Change the limits of integration:

When $x = 0$, $u = 1 + 4(0) = 1$.

When $x = 1$, $u = 1 + 4(1) = 5$.

The integral becomes:

$\int_1^5 u^{1/2} \frac{1}{4} du = \frac{1}{4} \int_1^5 u^{1/2} du$

$= \frac{1}{4} \left[ \frac{u^{3/2}}{3/2} \right]_1^5$

$= \frac{1}{4} \left[ \frac{2}{3} u^{3/2} \right]_1^5$

$= \frac{1}{6} [u^{3/2}]_1^5$

$= \frac{1}{6} (5^{3/2} - 1^{3/2})$

$= \frac{1}{6} (5\sqrt{5} - 1)$

Now, substitute these results back into the expression for $I$:

$I = \frac{1}{2} \left[ 1 + \frac{1}{6} (5\sqrt{5} - 1) \right]$

$I = \frac{1}{2} \left[ \frac{6}{6} + \frac{5\sqrt{5} - 1}{6} \right]$

$I = \frac{1}{2} \left[ \frac{6 + 5\sqrt{5} - 1}{6} \right]$

$I = \frac{1}{2} \left[ \frac{5 + 5\sqrt{5}}{6} \right]$

$I = \frac{5 + 5\sqrt{5}}{12}$

$I = \frac{5}{12} (1 + \sqrt{5})$

Comparing this result with the given options, it matches option C.