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Use the table of standard integrals to find the exact value of $$\int_0^2 \frac{dx}{\sqrt{16 - x^2}}.$$ (b) Find $$\frac{d}{dx} (x \sin^2 x).$$ (c) Evaluate $$\sum_{n=4}^7 (2n + 3).$$ (d) Let A be the point (-2, 7) and let B be the point (1, 5) - HSC - SSCE Mathematics Extension 1 - Question 1 - 2001 - Paper 1
Question 1
Use the table of standard integrals to find the exact value of $$\int_0^2 \frac{dx}{\sqrt{16 - x^2}}.$$ (b) Find $$\frac{d}{dx} (x \sin^2 x).$$ (c) Evaluate ... show full transcript
Worked Solution & Example Answer:Use the table of standard integrals to find the exact value of $$\int_0^2 \frac{dx}{\sqrt{16 - x^2}}.$$ (b) Find $$\frac{d}{dx} (x \sin^2 x).$$ (c) Evaluate $$\sum_{n=4}^7 (2n + 3).$$ (d) Let A be the point (-2, 7) and let B be the point (1, 5) - HSC - SSCE Mathematics Extension 1 - Question 1 - 2001 - Paper 1
Step 1
Use the table of standard integrals to find the exact value of $$\int_0^2 \frac{dx}{\sqrt{16 - x^2}}.$$
Answer
To evaluate the integral , we recognize it as a standard integral form related to inverse trigonometric functions. Using the substitution ( x = 4 \sin \theta ), we have ( dx = 4 \cos \theta d\theta ) and the limits change from 0 to 2 into 0 to ( \frac{\pi}{6} ). Thus, we transform the integral:
Step 2
Find $$\frac{d}{dx} (x \sin^2 x)$$.
Answer
To find the derivative of (x \sin^2 x), we will use the product rule, which states that ( (uv)' = u'v + uv'). Here, let (u = x) and (v = \sin^2 x).
[\frac{d}{dx}(x \sin^2 x) = \sin^2 x + x \cdot \frac{d}{dx}(\sin^2 x) = \sin^2 x + x \cdot 2\sin x \cos x = \sin^2 x + x \sin(2x).]
Step 3
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.
Answer
To find point P which divides AB externally in the ratio 1:2, we use the 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 A = (x_1, y_1) = (-2, 7), B = (x_2, y_2) = (1, 5), m = 1, and n = 2.
Calculating, we get: [ P = \left( \frac{1(1) - 2(-2)}{1 - 2}, \frac{1(5) - 2(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.
Answer
To determine if (x + 3) is a factor, we can use synthetic division or evaluate the polynomial at (x = -3) (the root). If the polynomial evaluates to zero, then it is a factor:
[f(-3) = (-3)^3 - 5(-3) + 12 = -27 + 15 + 12 = 0.]
Since (f(-3) = 0), (x + 3) is indeed a factor of the polynomial (x^3 - 5x + 12).
Step 6
Use the substitution \(u = 1 + x\) to evaluate $$15 \int_{-1}^1 \frac{\sqrt{1 + x}}{x} dx.$$
Answer
First, we substitute (u = 1 + x), so when (x = -1), (u = 0) and when (x = 1), (u = 2). The integral becomes: [15 \int_0^2 \frac{\sqrt{u}}{u - 1} du.]
This requires us to break the integral at the point of discontinuity (given by (u=1)). The integral can be computed by splitting it:
[= 15 \left( \int_0^1 \frac{\sqrt{u}}{u - 1} du + \int_1^2 \frac{\sqrt{u}}{u - 1} du \right).]
The evaluations will require further simplification and methods such as integration by parts or numerical techniques for the final calculation.