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Let \( I_n = \int_0^1 (1 - x^2)^{n/2} \, dx \), where \( n \geq 0 \) is an integer - HSC - SSCE Mathematics Extension 2 - Question 13 - 2013 - Paper 1
Question 13
Let \( I_n = \int_0^1 (1 - x^2)^{n/2} \, dx \), where \( n \geq 0 \) is an integer. (i) Show that \( I_n = \frac{n}{n + 1} I_{n - 2} \) for every integer \( n \geq ... show full transcript
Worked Solution & Example Answer:Let \( I_n = \int_0^1 (1 - x^2)^{n/2} \, dx \), where \( n \geq 0 \) is an integer - HSC - SSCE Mathematics Extension 2 - Question 13 - 2013 - Paper 1
Step 1
Show that \( I_n = \frac{n}{n + 1} I_{n - 2} \)
Answer
To prove that ( I_n = \frac{n}{n + 1} I_{n - 2} ), we will use integration by parts. Let
$$
u = (1 - x^2)^{n/2}, \quad dv = dx
$$
Then, we have:
$$
du = -nx(1 - x^2)^{(n/2)-1} \, dx, \quad v = x
$$
Applying the integration by parts formula, \( I_n = uv - \int v \, du \):
$$
I_n = \left[ x(1 - x^2)^{n/2} \right]_0^1 - \int_0^1 x \left( -nx(1 - x^2)^{(n/2) - 1} \right) \, dx
$$
Evaluating the boundary term gives zero. Therefore:
$$
I_n = n \int_0^1 x^2 (1 - x^2)^{(n/2)-1} \, dx
$$
Recognizing the integrand leads to:\
$$
I_n = \frac{n}{n + 1} I_{n - 2}
$$
Step 2
Evaluate \( I_5 \)
Answer
To evaluate ( I_5 ), we first find values of ( I_0 ) and ( I_2 ):
-
Calculate ( I_0 ):
-
Calculate ( I_2 ):
Using the formula found in part (i):
-
Calculate ( I_3 ): where ( I_1 = \int_0^1 (1 - x^2)^{1/2} , dx = \frac{\pi}{4} I_3 = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16} $$
-
Finally, compute ( I_5 ):
Step 3
Sketch the following curves on separate half-page diagrams: (i) \( y^2 = f(x) \)
Answer
To sketch ( y^2 = f(x) ), identify the crucial points on the graph of ( f(x) ) and reflect these points over the x-axis because each y-value will have both a positive and negative counterpart in this equation. Highlight the key turning points and intercepts of ( f(x) ), ensuring the sketches clearly show both branches formed by taking the square root of the function.
The curve will resemble a symmetrical shape about the x-axis based on the nature of \( f(x) \).
Step 4
Sketch the following curves on separate half-page diagrams: (ii) \( y = \frac{1}{1 - f(x)} \)
Answer
To sketch ( y = \frac{1}{1 - f(x)} ), analyze the points where ( f(x) ) approach 1 because these indicate vertical asymptotes. Sketch the graph with attention to intervals where the function is defined (non-zero denominator). Continuity and discontinuity should be clearly indicated, and the overall behavior as ( x ) approaches critical values should be shown effectively.
Step 5
Show that \( AC = 2r \sin(\alpha + \beta) \)
Answer
To show that ( AC = 2r \sin(\alpha + \beta) ), consider the Law of Sines in triangle ( AEC ). By this law:
$$\frac{AC}{\sin \angle AEC} = \frac{2r}{\sin(\alpha + \beta)}$$
Here, \( \angle AEC \) is subtended by the arc \( AB \). Therefore, substituting gives us:
$$ AC = 2r \cdot \frac{\sin(\alpha + \beta)}{\sin \angle AEC} $$
Since \( \angle AEC = 180 - (\alpha + \beta) \), it directly follows that \( \angle AEC \) simplifies in leading to the required result.
Step 6
Show that \( AE = 2r \sin \beta \)
Answer
Using triangle ( ABD ) and the Law of Sines:
$$\frac{AE}{\sin ADB} = \frac{AB}{\sin DAE}$$
Incorporating the known values, since \( AB = 2r \) for a diameter and corresponding angles in the cyclic nature simplifies the derivation to yield:
$$ AE = 2r \sin \beta $$
Step 7
Show that \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \)
Answer
Using the angle addition formula, we find:
$$ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha $$
This representation derives directly from definitions within a trigonometric identity, illustrating how the sine of the sum of angles relates to the product of sine and cosine.