SSCE HSC Mathematics Extension 1 Integrals involving trigonometric substitution: 1. (a) Use the table of standard

1. (a) 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

Question 1

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1. (a) Use the table of standard integrals to find the exact value of \[ \int_{0}^{2} \frac{dx}{\sqrt{16 - x^2}}. \] 1. (b) Find \[ \frac{d}{dx}(x \s... show full transcript

Worked Solution & Example Answer:1. (a) 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

Step 1

Use the table of standard integrals to find the exact value of $ \int_{0}^{2} \frac{dx}{\sqrt{16 - x^2}} $

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Answer

To solve this integral, we recognize that it is of the form ( \int \frac{dx}{\sqrt{a^2 - x^2}} ), which evaluates to ( \sin^{-1}(\frac{x}{a}) + C ). Here, (a = 4). Thus, we have:

[ \int_{0}^{2} \frac{dx}{\sqrt{16 - x^2}} = \left[ \sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{2} = \sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}. ]

Step 2

Find $ \frac{d}{dx}(x \sin^2 x) $

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Answer

Using the product rule for differentiation, we have:

[ \frac{d}{dx}(x \sin^2 x) = \sin^2 x + x \cdot 2\sin x \cos x = \sin^2 x + x \sin(2x). ]

Step 3

Evaluate $ \sum_{n=4}^{7}(2n + 3) $

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Answer

Calculating the sum term by 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)

Therefore, the total sum is:

[ 11 + 13 + 15 + 17 = 56. ]

Step 4

Let A be the point (-2, 7) and let B be the point (1, 5). Find the coordinates of the point P which divides the interval AB externally in the ratio 1:2.

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Answer

Using the section formula for external division:

[ P = \left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n} \right) ]

where (m = 1), (n = 2), (A(-2, 7)), and (B(1, 5)):

[ P = \left( \frac{1 \cdot 1 - 2 \cdot (-2)}{1 - 2}, \frac{1 \cdot 5 - 2 \cdot 7}{1 - 2} \right) = \left( \frac{1 + 4}{-1}, \frac{5 - 14}{-1} \right) = \left( -5, 9 \right). ]

Step 5

Is $x + 3$ a factor of $x^3 - 5x + 12$? Give reasons for your answer.

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Answer

To determine if (x + 3) is a factor, we evaluate the polynomial at (x = -3):

[ f(-3) = (-3)^3 - 5(-3) + 12 = -27 + 15 + 12 = 0. ]

Since the result is zero, (x + 3) is indeed a factor according to the Factor Theorem.

Step 6

Use the substitution $u = 1 + x$ to evaluate $15 \int_{-1}^{1} \frac{1}{\sqrt{1 + x}} dx$.

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Answer

With the substitution (u = 1 + x), we have (du = dx), and the limits change to (u = 0) when (x = -1) and (u = 2) when (x = 1). Thus, the integral becomes:

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

1. (a) 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

Worked Solution & Example Answer:1. (a) 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

Use the table of standard integrals to find the exact value of \( \int_{0}^{2} \frac{dx}{\sqrt{16 - x^2}} \)

Find \( \frac{d}{dx}(x \sin^2 x) \)

Evaluate \( \sum_{n=4}^{7}(2n + 3) \)

Let A be the point (-2, 7) and let B be the point (1, 5). Find the coordinates of the point P which divides the interval AB externally in the ratio 1:2.

Is \(x + 3\) a factor of \(x^3 - 5x + 12\)? Give reasons for your answer.

Use the substitution \(u = 1 + x\) to evaluate \(15 \int_{-1}^{1} \frac{1}{\sqrt{1 + x}} dx\).

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