Use the table of standard integrals to find the exact value of $$ \int_{0}^{2} \frac{dx}{\sqrt{16 - x^{2}}} - HSC - SSCE Mathematics Extension 1 - Question 1 - 2001 - Paper 1

We use the product rule for differentiation:

$\frac{d}{dx}(u \cdot v) = u'v + uv'$,

where $u = x$ and $v = \sin^{2} x$. Calculating:

• $u' = 1$,
• $v' = 2\sin{x}\cos{x} = \sin{2x}$,

So,

$\frac{d}{dx}(x \sin^{2} x) = 1 \cdot \sin^{2} x + x \cdot \sin{2x}.$

Calculating each term:

• For n=4: $2(4) + 3 = 11$,
• For n=5: $2(5) + 3 = 13$,
• For n=6: $2(6) + 3 = 15$,
• For n=7: $2(7) + 3 = 17$.

The total sum is:

$11 + 13 + 15 + 17 = 56.$

Using the section formula for external division:

$P = \left(\frac{mx_{2} - nx_{1}}{m - n}, \frac{my_{2} - ny_{1}}{m - n}\right)$, where m = 1 and n = 2:

$P = \left(\frac{1 \cdot 1 - 2 \cdot (-2)}{1 - 2}, \frac{1 \cdot 5 - 2 \cdot 7}{1 - 2}\right) = (3, -9).$

To determine if $x + 3$ is a factor, we apply the Factor Theorem. We evaluate the polynomial at $x = -3$:

$(-3)^{3} - 5(-3) + 12 = -27 + 15 + 12 = 0.$

Since the result is zero, $x + 3$ is indeed a factor.

Using the substitution $u = 1 + x$, we find:

• When $x = -1$, $u = 0$;
• When $x = 1$, $u = 2$.

Thus, we convert the integral:

$15 \int_{0}^{2} \frac{1}{\sqrt{u}} du = 15[2\sqrt{u}]_{0}^{2} = 15(2 \cdot \sqrt{2} - 0) = 30 \sqrt{2}.$