SSCE HSC Mathematics Extension 1 Integration involving inverse trigonometric functions: Use the table of standard integr

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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Use the table of standard integrals to find the exact value of $$ \int_0^2 \frac{dx}{\sqrt{16 - x^2}}. $$ Find $$ \frac{d}{dx}(\sin^2 x). $$ Evaluate $$ \sum_{n=4}... show full transcript

Worked Solution & Example Answer: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 evaluate the integral

\int_0^2 \frac{dx}{\sqrt{16 - x^2}},

we can use the standard integral result:

\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C.

Here, ( a = 4 ).

Thus:

\int_0^2 \frac{dx}{\sqrt{16 - x^2}} = \left[\arcsin\left(\frac{x}{4}\right)\right]_0^2 = \arcsin\left(\frac{2}{4}\right) - \arcsin(0) = \arcsin\left(\frac{1}{2}\right) - 0 = \frac{\pi}{6}.

Step 2

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

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Answer

Using the chain rule:

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

Thus, the derivative is

\frac{d}{dx}(\sin^2 x) = 2\sin x \cos x = \sin(2x).

Step 3

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

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Answer

Calculating the individual terms:

For ( n=4 ): ( 2(4) + 3 = 8 + 3 = 11 )
For ( n=5 ): ( 2(5) + 3 = 10 + 3 = 13 )
For ( n=6 ): ( 2(6) + 3 = 12 + 3 = 15 )
For ( n=7 ): ( 2(7) + 3 = 14 + 3 = 17 )

Adding these values gives:

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: Given points A(-2, 7) and B(1, 5) and the ratio of division 1:2,

P = \left(\frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n}\right) = \left(\frac{1 \cdot 1 - 2 \cdot (-2)}{1 - 2}, \frac{1 \cdot 5 - 2 \cdot 7}{1 - 2}\right)

Calculating:

P = \left(\frac{1 + 4}{-1}, \frac{5 - 14}{-1}\right) = \left(-5, 9\right).

Step 5

Is x + 3 a factor of x³ - 5x + 12? Give reasons for your answer.

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Answer

To determine if ( x + 3 ) is a factor of ( x^3 - 5x + 12 ), we can use synthetic division or substitute ( x = -3 ):

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

Since the result is 0, ( x + 3 ) is indeed a factor of ( x^3 - 5x + 12 ).

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

Performing the substitution ( u = 1 + x ) transforms the limits as follows:

When ( x = -1 ), ( u = 0 )
When ( x = 1 ), ( u = 2 )

Thus:

15 \int_{0}^{2} \frac{1}{\sqrt{u}} du

This integral evaluates to:

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

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

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 x³ - 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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