Long Question 1

Posted on April 22, 2020 by user

Question 1:

The above diagram shows triangle OAB. The straight line AP intersects the straight line OQ at R. It is given that

O P = \frac{1}{4} O B, A Q = \frac{1}{4} A B, \vec{O P} = 4 \underset{˜}{b} and \vec{O A} = 8 \underset{˜}{a} .

(a) Express in terms of

\underset{˜}{a} and/ or \underset{˜}{b} :

\begin{array}{l} (i) \vec{A P} \\ (ii) \vec{O Q} \end{array}

(b)(i) Given that

\vec{A R} = h \vec{A P}

, state

\vec{A R}

in terms of h,

\underset{˜}{a} and \underset{˜}{b} .

(ii) Given that

\vec{R Q} = k \vec{O Q},

state in terms of k,

\underset{˜}{a} and \underset{˜}{b} .

(c) Using

\vec{A Q} = \vec{A R} + \vec{R Q},

find the value of h and of k.

Solution:
(a)(i)

\begin{array}{l} \vec{A P} = \vec{A O} + \vec{O P} \\ \vec{A P} = - \vec{O A} + \vec{O P} \\ \vec{A P} = - 8 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}

(a)(ii)

\begin{array}{l} \vec{O Q} = \vec{O A} + \vec{A Q} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} \vec{A B} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (\vec{A O} + \vec{O B}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 \vec{O P}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 (4 \underset{˜}{b})) \\ \vec{O Q} = 8 \underset{˜}{a} - 2 \underset{˜}{a} + 4 \underset{˜}{b} \\ \vec{O Q} = 6 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}

(b)(i)

\begin{array}{l} \vec{A R} = h \vec{A P} \\ \vec{A R} = h (- 8 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{A R} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} \end{array}

(b)(ii)

\begin{array}{l} \vec{R Q} = k \vec{O Q} \\ \vec{R Q} = k (6 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{R Q} = 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \end{array}

(c)

\begin{array}{l} \vec{A Q} = \vec{A R} + \vec{R Q} \\ \vec{A Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + (6 k \underset{˜}{a} + 4 k \underset{˜}{b}) \\ \vec{A O} + \vec{O Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \\ - 8 \underset{˜}{a} + 6 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 6 k \underset{˜}{a} + 4 h \underset{˜}{b} + 4 k \underset{˜}{b} \\ - 2 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 6 k \underset{˜}{a} + 4 h \underset{˜}{b} + 4 k \underset{˜}{b} \\ - 2 = - 8 h + 6 k \\ - 1 = - 4 h + 3 k \to (1) \\ 4 = 4 h + 4 k \\ 1 = h + k \\ k = 1 - h \to (2) \\ Substitute (2) into (1), \\ - 1 = - 4 h + 3 (1 - h) \\ - 1 = - 4 h + 3 - 3 h \\ - 4 = - 7 h \\ h = \frac{4}{7} \\ From (2), \\ k = 1 - \frac{4}{7} = \frac{3}{7} \end{array}

Long Question 3

Posted on April 22, 2020 by user

Question 3:

In diagram below, PQRS is a quadrilateral. PTS and TUR are straight lines.

It is given that

\vec{P Q} = 20 \underset{˜}{x}, \vec{P T} = 8 \underset{˜}{y}, \vec{S R} = 25 \underset{˜}{x} - 24 \underset{˜}{y}, \vec{P T} = \frac{1}{4} \vec{P S} and \vec{T U} = \frac{3}{5} \vec{T R}

(a) Express in terms of

\underset{˜}{x}

and/or

\underset{˜}{y}

(i) $\vec{Q S}$
(ii) $\vec{T R}$
(b) Show that the points Q, U and S are collinear.

(c) If

| \underset{˜}{x} |

= 2 and

| \underset{˜}{y} |

= 3, find

| \vec{Q S} |

Solution:
(a)(i)

\begin{array}{l} \vec{Q S} = \vec{Q P} + \vec{P S} \\ \vec{Q S} = - 20 \underset{˜}{x} + 32 \underset{˜}{y} \leftarrow \begin{array}{l} Given \vec{P T} = \frac{1}{4} \vec{P S} \\ \vec{P S} = 4 \vec{P T} = 4 (8 \underset{˜}{y}) = 32 \underset{˜}{y} \end{array} \end{array}

(a)(ii)

\begin{array}{l} \vec{T R} = \vec{T S} + \vec{S R} \\ \vec{T R} = \frac{3}{4} \vec{P S} + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = \frac{3}{4} (32 \underset{˜}{y}) + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = 24 \underset{˜}{y} + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = 25 \underset{˜}{x} \end{array}

(b)

\begin{array}{l} \vec{Q U} = \vec{Q P} + \vec{P T} + \vec{T U} \\ \vec{Q U} = - 20 \underset{˜}{x} + 8 \underset{˜}{y} + \frac{3}{5} (25 \underset{˜}{x}) \leftarrow \begin{array}{l} Given \\ \vec{T U} = \frac{3}{5} \vec{T R} \end{array} \\ \vec{Q U} = - 20 \underset{˜}{x} + 8 \underset{˜}{y} + 15 \underset{˜}{x} \\ \vec{Q U} = - 5 \underset{˜}{x} + 8 \underset{˜}{y} \\ From (a)(i) \vec{Q S} = - 20 \underset{˜}{x} + 32 \underset{˜}{y} \\ \frac{\vec{Q S}}{\vec{Q U}} = \frac{- 20 \underset{˜}{x} + 32 \underset{˜}{y}}{- 5 \underset{˜}{x} + 8 \underset{˜}{y}} \\ \frac{\vec{Q S}}{\vec{Q U}} = \frac{4 (- 5 \underset{˜}{x} + 8 \underset{˜}{y})}{(- 5 \underset{˜}{x} + 8 \underset{˜}{y})} \\ \frac{\vec{Q S}}{\vec{Q U}} = 4 \\ \vec{Q S} = 4 \vec{Q U} \\ ∴ Q, U and S are collinear . \end{array}

(c)

\begin{array}{l} \vec{P S} = 32 \underset{˜}{y} \\ | \vec{P S} | = 32 | \underset{˜}{y} | \\ | \vec{P S} | = 32 \times 3 = 96 \\ \vec{P Q} = 20 \underset{˜}{x} \\ | \vec{P Q} | = 20 | \underset{˜}{x} | \\ | \vec{P Q} | = 20 \times 2 = 40 \\ ∴ | \vec{Q S} | = \sqrt{96^{2} + 40^{2}} \\ | \vec{Q S} | = 104 \end{array}

Integration as the Inverse of Differentiation

Posted on April 22, 2020 by user

3.4c Integration as the Inverse of Differentiation

Example:

\begin{array}{l} Shows that \frac{d}{d x} [\frac{2 x + 5}{x^{2} - 3}] = \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} \\ Hence, find the value of \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x \end{array}

Solution:

\begin{array}{l} \frac{d}{d x} [\frac{2 x + 5}{x^{2} - 3}] = \frac{(x^{2} - 3) (2) - (2 x + 5) (2 x)}{{(x^{2} - 3)}^{2}} \\ = \frac{2 x^{2} - 6 - 4 x^{2} - 10 x}{{(x^{2} - 3)}^{2}} \\ = \frac{- 2 x^{2} - 10 x - 6}{{(x^{2} - 3)}^{2}} \\ = \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} \\ \int_{0}^{2} \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = {[\frac{2 x + 5}{x^{2} - 3}]}_{0}^{2} \\ - 2 \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = {[\frac{2 x + 5}{x^{2} - 3}]}_{0}^{2} \\ \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = - \frac{1}{2} [(\frac{2 (2) + 5}{2^{2} - 3}) - (\frac{2 (0) + 5}{0^{2} - 3})] \\ = - \frac{1}{2} [9 - (- \frac{5}{3})] \\ = - \frac{1}{2} \times \frac{32}{3} \\ = - \frac{16}{3} \\ = - 5 \frac{1}{3} \end{array}

3.6 Integration as the Summation of Volumes

Posted on April 22, 2020 by user

3.6 Integration as the Summation of Volumes

(1).

The volume of the solid generated when the region enclosed by the curve y = f(x), the x-axis, the line x = a and the line x = b is revolved through 360° about the x-axis is given by

V_{x} = π \int_{a}^{b} y^{2} d x

(2).

The volume of the solid generated when the region enclosed by the curve x = f(y), the y-axis, the line y = a and the line y = b is revolved through 360° about the y-axis is given by

V_{y} = π \int_{a}^{b} x^{2} d y

Long Question 1 & 2

Posted on April 22, 2020 by user

Question 1:

A curve with gradient function

5 x - \frac{5}{x^{2}}

has a turning point at (m, 9).

(a) Find the value of m.

(b) Determine whether the turning point is a maximum or a minimum point.
(c) Find the equation of the curve.

Solution:
(a)

\begin{array}{l} \frac{d y}{d x} = 5 x - \frac{5}{x^{2}} \\ At turning point (m, 9), \frac{d y}{d x} = 0. \\ 5 m - \frac{5}{m^{2}} = 0 \\ \frac{5}{m^{2}} = 5 m \\ m^{3} = 1 \\ m = 1 \end{array}

(b)

\begin{array}{l} \frac{d y}{d x} = 5 x - \frac{5}{x^{2}} = 5 x - 5 x^{- 2} \\ \frac{d^{2} y}{d x^{2}} = 5 + \frac{10}{x^{3}} \\ When x = 1, \frac{d^{2} y}{d x^{2}} = 15 (> 0) \\ Thus, (1, 9) is a minimum point . \end{array}

(c)

\begin{array}{l} y = \int (5 x - 5 x^{- 2}) d x \\ y = \frac{5 x^{2}}{2} + \frac{5}{x} + c \\ At turning point (1, 9), x = 1 and y = 9. \\ 9 = \frac{5 {(1)}^{2}}{2} + \frac{5}{1} + c \\ c = \frac{3}{2} \\ Equation of the curve: y = \frac{5 x^{2}}{2} + \frac{5}{x} + \frac{3}{2} \end{array}

Question 2:

A curve has a gradient function kx²– 7x, where k is a constant. The tangent to the curve at the point (1, 3 ) is parallel to the straight line y + x– 4 = 0.

Find

(a) the value of k,

(b) the equation of the curve.

Solution:
(a)

y + x – 4 = 0

y = – x + 4

m = –1

f ’(x) = kx² – 7x

Given tangent to the curve at the point (1, 3) parallel to the straight line

k (1)² – 7 (1) = –1

k – 7 = –1

k = 6

(b)

\begin{array}{l} f' (x) = 6 x^{2} - 7 x \\ f (x) = \int (6 x^{2} - 7 x) d x \\ f (x) = \frac{6 x^{3}}{3} - \frac{7 x^{2}}{2} + c \\ 3 = 2 {(1)}^{3} - \frac{7 {(1)}^{2}}{2} + c at point (1, 3) \\ c = \frac{9}{2} \\ ∴ f (x) = 2 x^{3} - \frac{7 x^{2}}{2} + \frac{9}{2} \end{array}

Long Question 5

Posted on April 22, 2020 by user

Question 5:
In Diagram below, the straight line WY is normal to the curve

y = \frac{1}{2} x^{2} + 1

at B (2, 4). The straight line BQ is parallel to the y–axis.

Find
(a) the value of t,
(b) the area of the shaded region,
(c) the volume generated, in terms of π, when the region bounded by the curve, the y–axis
and the straight line y = 4 is revolved through 360° about the y-axis.

Solution:
(a)

\begin{array}{l} y = \frac{1}{2} x^{2} + 1 \\ Gradient of tangent, \frac{d y}{d x} = 2 (\frac{1}{2} x) = x \\ At point B, \frac{d y}{d x} = 2 \\ Gradient of normal, m_{2} = - \frac{1}{2} \\ \frac{4 - 0}{2 - t} = - \frac{1}{2} \\ 8 = - 2 + t \\ t = 10 \end{array}

(b)

\begin{array}{l} Area of the shaded region \\ = Area under the curve + Area of triangle B Q Y \\ = \int_{0}^{2} (\frac{1}{2} x^{2} + 1) d x + \frac{1}{2} (10 - 2) (4) \\ = {[\frac{x^{3}}{6} + x]}_{0}^{2} + 16 \\ = [\frac{8}{6} + 2] - 0 + 16 \\ = 19 \frac{1}{3} {unit}^{2} \end{array}

(c)

\begin{array}{l} At y - axis, x = 0, y = \frac{1}{2} (0) + 1 = 1 \\ y = \frac{1}{2} x^{2} + 1, x^{2} = 2 y - 2 \\ Volume generated \\ = π \int x^{2} d y \\ = π \int_{1}^{4} (2 y - 2) d y \\ = π {[y^{2} - 2 y]}_{1}^{4} \\ = π [(16 - 8) - (1 - 2)] \\ = 9 π {unit}^{3} \end{array}

Short Question 1

Posted on April 22, 2020 by user

Question 1:

\begin{array}{l} Find the integral of each of the following . \\ (a) \int (\frac{3}{x^{2}} - \frac{5}{2 x^{3}} + 2) d x \\ (b) \int x^{2} (x^{5} + 2 x) d x \\ (c) \int \frac{3 x^{4} + 2 x}{x^{3}} d x \\ (d) \int \frac{(7 + x) (7 - x)}{x^{4}} d x \\ (e) \int {(5 x - 1)}^{3} d x \\ (f) \int \frac{3}{{(4 x + 7)}^{8}} d x \end{array}

Solution:
(a)

\begin{array}{l} \int (\frac{3}{x^{2}} - \frac{5}{2 x^{3}} + 2) d x \\ = \int (3 x^{- 2} - \frac{5 x^{- 3}}{2} + 2) d x \\ = - 3 x^{- 1} + \frac{5 x^{- 2}}{4} + 2 x + c \\ = - \frac{3}{x} + \frac{5}{4 x^{2}} + 2 x + c \end{array}

(b)

\begin{array}{l} \int x^{2} (x^{5} + 2 x) d x \\ = \int (x^{7} + 2 x^{3}) d x \\ = \frac{x^{8}}{8} + \frac{2 x^{4}}{4} + c \\ = \frac{x^{8}}{8} + \frac{x^{4}}{2} + c \end{array}

(c)

\begin{array}{l} \int \frac{3 x^{4} + 2 x}{x^{3}} d x \\ = \int (\frac{3 x^{4}}{x^{3}} + \frac{2 x}{x^{3}}) d x = \int (3 x + 2 x^{- 2}) d x \\ = \frac{3 x^{2}}{2} - \frac{2}{x} + c \end{array}

(d)

\begin{array}{l} \int \frac{(7 + x) (7 - x)}{x^{4}} d x = \int (\frac{49 - x^{2}}{x^{4}}) d x \\ = \int (\frac{49}{x^{4}} - \frac{1}{x^{2}}) d x = \int (49 x^{- 4} - x^{- 2}) d x \\ = \frac{49 x^{- 3}}{- 3} + \frac{1}{x} + c \\ = - \frac{49}{3 x^{3}} + \frac{1}{x} + c \end{array}

(e)

\begin{array}{l} \int {(5 x - 1)}^{3} d x \\ = \frac{{(5 x - 1)}^{4}}{(4) (5)} + c \\ = \frac{1}{20} {(5 x - 1)}^{4} + c \end{array}

(f)

\begin{array}{l} \int \frac{3}{{(4 x + 7)}^{8}} d x = \int 3 {(4 x + 7)}^{- 8} d x \\ = \frac{3 {(4 x + 7)}^{- 7}}{(- 7) (4)} + c \\ = - \frac{3}{28 {(4 x + 7)}^{7}} + c \end{array}

Short Question 5 – 7

Posted on April 22, 2020 by user

Question 5:

\begin{array}{l} Given \int (6 x^{2} + 1) d x = m x^{3} + x + c, \\ where m and c are constants, find \\ (a) the value of m . \\ (b) the value of c if \int (6 x^{2} + 1) d x = 13 when x = 1. \end{array}

Solution:
(a)

\begin{array}{l} \int (6 x^{2} + 1) d x = m x^{3} + x + c \\ \frac{6 x^{3}}{3} + x + c = m x^{3} + x + c \\ 2 x^{3} + x + c = m x^{3} + x + c \\ Compare the both sides, \\ ∴ m = 2 \end{array}

(b)

\begin{array}{l} \int (6 x^{2} + 1) d x = 13 when x = 1. \\ 2 {(1)}^{3} + 1 + c = 13 \\ 3 + c = 13 \\ c = 10 \end{array}

Question 6:

\begin{array}{l} It is given that \int_{5}^{k} g (x) d x = 6, and \int_{5}^{k} [g (x) + 2] d x = 14, \\ find the value of k . \end{array}

Solution:

\begin{array}{l} \int_{5}^{k} [g (x) + 2] d x = 14 \\ \int_{5}^{k} g (x) d x + \int_{5}^{k} 2 d x = 14 \\ 6 + {[2 x]}_{5}^{k} = 14 \\ 2 (k - 5) = 8 \\ k - 5 = 4 \\ k = 9 \end{array}

Question 7:

Given \int_{k}^{2} (4 x + 7) d x = 28, calculate the possible value of k .

Solution:

\begin{array}{l} \int_{k}^{2} (4 x + 7) d x = 28 \\ {[2 x^{2} + 7 x]}_{k}^{2} = 28 \\ 8 + 14 - (2 k^{2} + 7 k) = 28 \\ 22 - 2 k^{2} - 7 k = 28 \\ 2 k^{2} + 7 k + 6 = 0 \\ (2 k + 3) (k + 2) = 0 \\ k = - \frac{3}{2} or k = - 2 \end{array}

Short Question 8 – 10

Posted on April 22, 2020 by user

Question 8:

Given y = \frac{5 x}{x^{2} + 1} and \frac{d y}{d x} = g (x), find the value of \int_{0}^{3} 2 g (x) d x .

Solution:

\begin{array}{l} Since \frac{d y}{d x} = g (x), thus y = \int g (x) d x \\ \int_{0}^{3} 2 g (x) d x = 2 \int_{0}^{3} g (x) d x \\ = 2 {[y]}_{0}^{3} \\ = 2 {[\frac{5 x}{x^{2} + 1}]}_{0}^{3} \\ = 2 [\frac{5 (3)}{3^{2} + 1} - 0] \\ = 2 (\frac{15}{10}) \\ = 3 \end{array}

Question 9:

Find \int_{5}^{k} (x + 1) d x, in terms of k .

Solution:

\begin{array}{l} {\int_{5}^{k} (x + 1) d x = [\frac{x^{2}}{2} + x]}_{5}^{k} \\ = (\frac{k^{2}}{2} + k) - (\frac{5^{2}}{2} + 5) \\ = \frac{k^{2} + 2 k}{2} - \frac{35}{2} \\ = \frac{k^{2} + 2 k - 35}{2} \end{array}

Question 10:

\begin{array}{l} Given that = \int_{2}^{5} g (x) d x = - 2 . Find \\ (a) the value of \int_{5}^{2} g (x) d x, \\ (b) the value of m if \int_{2}^{5} [g (x) + m (x)] d x = 19 \end{array}

Solution:

\begin{array}{l} (a) \int_{5}^{2} g (x) d x = - \int_{2}^{5} g (x) d x \\ = - (- 2) \\ = 2 \end{array}

\begin{array}{l} (b) \int_{2}^{5} [g (x) + m (x)] d x = 19 \\ \int_{2}^{5} g (x) d x + m \int_{2}^{5} x d x = 19 \\ - 2 + m {[\frac{x^{2}}{2}]}_{2}^{5} = 19 \\ \frac{m}{2} {[x^{2}]}_{2}^{5} = 21 \\ \frac{m}{2} [25 - 4] = 21 \\ 21 m = 42 \\ m = 2 \end{array}

Short Question 11 – 13

Posted on April 22, 2020 by user

Question 11:

\begin{array}{l} Given \int_{- 2}^{3} g (x) d x = 4, and \int_{- 2}^{3} h (x) d x = 9, find the value of \\ (a) \int_{- 2}^{3} 5 g (x) d x, \\ (b) m if \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \end{array}

Solution:
(a)

\begin{array}{l} \int_{- 2}^{3} 5 g (x) d x = 5 \int_{- 2}^{3} g (x) d x \\ = 5 \times 4 \\ = 20 \end{array}

(b)

\begin{array}{l} \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \\ \int_{- 2}^{3} g (x) d x + 3 \int_{- 2}^{3} h (x) d x + \int_{- 2}^{3} 4 m d x = 12 \\ 4 + 3 (9) + 4 m {[x]}_{- 2}^{3} = 12 \\ 4 m [3 - (- 2)] = - 19 \\ 20 m = - 19 \\ m = - \frac{19}{20} \end{array}

Question 12:

\begin{array}{l} (a) Find the value of \int_{- 1}^{1} {(3 x + 1)}^{3} d x . \\ (b) Evaluate \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x . \end{array}

Solution:

\begin{array}{l} a) {\int_{- 1}^{1} {(3 x + 1)}^{3} d x = [\frac{{(3 x + 1)}^{4}}{4 (3)}]}_{- 1}^{1} \\ = {[\frac{{(3 x + 1)}^{4}}{12}]}_{- 1}^{1} \\ = \frac{1}{12} [4^{4} - {(- 2)}^{4}] \\ = \frac{1}{12} (256 - 16) \\ = 20 \end{array}

\begin{array}{l} (b) \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x = \int_{3}^{4} \frac{1}{{(2 x - 4)}^{\frac{1}{2}}} d x \\ = \int_{3}^{4} {(2 x - 4)}^{- \frac{1}{2}} d x \\ = {[\frac{{(2 x - 4)}^{- \frac{1}{2} + 1}}{\frac{1}{2} (2)}]}_{3}^{4} \\ = {[\sqrt{2 x - 4}]}_{3}^{4} \\ = [\sqrt{2 (4) - 4} - \sqrt{2 (3) - 4}] \\ = 2 - \sqrt{2} \end{array}

Question 13:

\begin{array}{l} Given that y = \frac{x^{2}}{2 x - 1}, show that \\ \frac{d y}{d x} = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} . Hence, evaluate \int_{- 2}^{2} \frac{x (x - 1)}{4 {(2 x - 1)}^{2}} d x . \end{array}

Solution:

\begin{array}{l} y = \frac{x^{2}}{2 x - 1} \\ \frac{d y}{d x} = \frac{(2 x - 1) (2 x) - x (2)}{{(2 x - 1)}^{2}} \\ = \frac{4 x^{2} - 2 x - 2 x^{2}}{{(2 x - 1)}^{2}} \\ = \frac{2 x^{2} - 2 x}{{(2 x - 1)}^{2}} \\ = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} (shown) \\ \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{8} \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{4} \int_{- 2}^{2} \frac{x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} [(\frac{2^{2}}{2 (2) - 1}) - (\frac{{(- 2)}^{2}}{2 (- 2) - 1})] \\ = \frac{1}{8} [(\frac{4}{3}) - (\frac{4}{- 5})] \\ = \frac{1}{8} (\frac{32}{15}) \\ = \frac{4}{15} \end{array}