Let $I_n = \int_0^1 x^n \sqrt{1-x^2} \, dx$, for $n = 0, 1, 2, \ldots$ (i) Find the value of $I_1$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2017 - Paper 1

We use integration by parts to derive the recursive relation:

Letting $u = x^n$ and $dv = \sqrt{1-x^2} \, dx$, we have:

$du = n x^{n-1} \, dx, \quad v = -\frac{1}{2} (1-x^2)^{3/2}.$

Therefore, by integration by parts:

$I_n = uv \bigg|_0^1 - \int_0^1 v \, du = 0 - \int_0^1 -\frac{1}{2}(1-x^2)^{3/2} n x^{n-1} \, dx.$

This gives:

$I_n = \frac{n}{2} \int_0^1 x^{n-1} (1-x^2)^{3/2} \, dx.$

We can simplify the integral using a substitution and relate it back to $I_{n-2}$, resulting in:

$I_n = \frac{(n-1)}{n+2} I_{n-2}.$

Using the recursion established:

$I_5 = \frac{(5-1)}{(5+2)} I_{3} = \frac{4}{7} I_{3}.$

Continuing, we find:

$I_3 = \frac{(3-1)}{(3+2)} I_{1} = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}.$

Putting these values back, we have:

$I_5 = \frac{4}{7} \cdot \frac{2}{15} = \frac{8}{105}.$

Thus, $I_5 = \frac{8}{105}$.

To find the equation of the tangent line:

1. We start with the curve $\sqrt{x + y} = \sqrt{a}$.

2. Differentiate both sides with respect to $x$:

$\frac{1}{2\sqrt{x+y}} \left(1 + \frac{dy}{dx}\right) = 0.$

At point $P(c, d)$, we substitute $x = c$ and $y = d$ to find:

$\frac{dy}{dx} = -\frac{\sqrt{c}}{\sqrt{d}}.$

3. Then the equation of the tangent at point $P(c,d)$, using point-slope form, yields:

$y - d = -\frac{\sqrt{c}}{\sqrt{d}} (x - c).$

Rearranging yields:

$y = -\frac{\sqrt{c}}{\sqrt{d}} x + K \, \text{, with constants determined.}$

Let $A$ and $B$ be the x and y-intercepts of the tangent line respectively:

• For $A$, when $y = 0$, the equation is solved for $x$,
• For $B$, when $x = 0$, it's solved for $y$.

Using the intercepts calculations,

$OA + OB = c + d.$

By the definition of the curve, $c + d = a$ follows directly:

Thus, we establish that:

$OA + OB = a.$