Review:

Question 1

The point \( A \) has coordinates \((-1, 6)\) and the point \( B \) has coordinates \( (7, 2) \).

(i) Find the equation of the perpendicular bisector of \( AB \), in the form \( y = mx + c \).

Midpoint of \( AB \):

\[ \text{Midpoint} = \left( \frac{-1 + 7}{2}, \frac{6 + 2}{2} \right) = (3, 4) \]

Gradient of \( AB \):

\[ m_{AB} = \frac{2 - 6}{7 - (-1)} = \frac{-4}{8} = -\frac{1}{2} \]

Gradient of the perpendicular bisector:

\[ m_{\text{bisector}} = -\frac{1}{m_{AB}} = 2 \]

Equation of the perpendicular bisector:

\[ y - 4 = 2(x - 3) \implies y = 2x - 2 \]

Answer: \( y = 2x - 2 \).

(ii) A point \( C \) on the perpendicular bisector has coordinates \( (p, q) \), and the distance \( OC = 2 \). Write two equations involving \( p \) and \( q \), and find the possible coordinates of \( C \).

From the equation of the perpendicular bisector:

\[ q = 2p - 2 \]

Using the distance formula:

\[ \sqrt{p^2 + q^2} = 2 \implies p^2 + q^2 = 4 \]

Substitute \( q = 2p - 2 \):

\[ p^2 + (2p - 2)^2 = 4 \]

Expand:

\[ p^2 + 4p^2 - 8p + 4 = 4 \implies 5p^2 - 8p = 0 \]

Factorize:

\[ p(5p - 8) = 0 \implies p = 0 \, \text{or} \, p = \frac{8}{5} \]

For \( p = 0 \):

\[ q = 2(0) - 2 = -2 \implies C = (0, -2) \]

For \( p = \frac{8}{5} \):

\[ q = 2\left(\frac{8}{5}\right) - 2 = \frac{6}{5} \implies C = \left(\frac{8}{5}, \frac{6}{5}\right) \]

Answer: \( C = (0, -2) \) or \( C = \left(\frac{8}{5}, \frac{6}{5}\right) \).

Question 2

A piece of wire of length \( 24 \, \text{cm} \) is bent to form the perimeter of a sector of a circle of radius \( r \, \text{cm} \).

(i) Show that the area of the sector, \( A \, \text{cm}^2 \), is given by \( A = 12r - r^2 \).

Perimeter of the sector:

\[ 2r + l = 24 \implies l = 24 - 2r \]

Area of the sector:

\[ A = \frac{1}{2} r l \implies A = \frac{1}{2} r (24 - 2r) \implies A = 12r - r^2 \]

Answer: \( A = 12r - r^2 \).

(ii) Express \( A \) in the form \( a - (r - b)^2 \), where \( a \) and \( b \) are constants.

Complete the square:

\[ A = 12r - r^2 = -(r^2 - 12r) = -( (r - 6)^2 - 36 ) = 36 - (r - 6)^2 \]

Answer: \( A = 36 - (r - 6)^2 \).

(iii) Given that \( r \) can vary, state the greatest value of \( A \) and find the corresponding angle of the sector.

The maximum value of \( A \) occurs when \( (r - 6)^2 = 0 \):

\[ A = 36 \]

At \( r = 6 \), the angle of the sector:

\[ l = 24 - 2r = 12, \quad \text{Angle} = \frac{l}{r} \cdot 360 = \frac{12}{6} \cdot 360 = 720^\circ \]

Answer: Maximum \( A = 36 \), angle = \( 720^\circ \).

Question 3

Trigonometric identities and equations.

(i) Show that \( 2 \tan^2 \theta \sin^2 \theta = 1 \) can be written as \( 2 \sin^4 \theta + \sin^2 \theta - 1 = 0 \).

Rewrite \( \tan^2 \theta \):

\[ 2 \tan^2 \theta \sin^2 \theta = 2 \frac{\sin^2 \theta}{\cos^2 \theta} \sin^2 \theta = 2 \frac{\sin^4 \theta}{\cos^2 \theta} \]

Multiply through by \( \cos^2 \theta \):

\[ 2 \sin^4 \theta + \sin^2 \theta - 1 = 0 \]

Answer: \( 2 \sin^4 \theta + \sin^2 \theta - 1 = 0 \).

(ii) Solve \( 2 \tan^2 \theta \sin^2 \theta = 1 \) for \( 0^\circ \leq \theta \leq 360^\circ \).

Let \( x = \sin^2 \theta \):

\[ 2x^2 + x - 1 = 0 \]

Using the quadratic formula:

\[ x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{9}}{4} = \frac{1}{2} \]

\( \sin^2 \theta = \frac{1}{2} \):

\[ \sin \theta = \pm \frac{\sqrt{2}}{2} \implies \theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ \]

Answer: \( \theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ \).

(iii) Show that \( \frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta} = \frac{1}{\sin^2 \theta - \cos^2 \theta} \).

Combine the fractions:

\[ \frac{\sin \theta (\sin \theta - \cos \theta) + \cos \theta (\sin \theta + \cos \theta)}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)} \]

Simplify the numerator:

\[ \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta + \sin \theta \cos \theta = \sin^2 \theta + \cos^2 \theta = 1 \]

Answer: \( \frac{1}{\sin^2 \theta - \cos^2 \theta} \).

Review

Question 4

(i) In an arithmetic progression, the sum of the first 10 terms is 400 and the sum of the next 10 terms is 1000. Find the common difference and the first term.

The formula for the sum of the first \( n \) terms of an arithmetic progression (AP) is:

\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

For the first 10 terms:

\[ 400 = \frac{10}{2} \left(2a + 9d\right) \implies 2a + 9d = 80 \tag{1} \]

For the next 10 terms (terms 11 to 20):

The total sum of the first 20 terms is:

\[ 1000 + 400 = 1400 \]

So, for 20 terms:

\[ 1400 = \frac{20}{2} \left(2a + 19d\right) \implies 2a + 19d = 140 \tag{2} \]

Subtract equation (1) from equation (2):

\[ (2a + 19d) - (2a + 9d) = 140 - 80 \]

\[ 10d = 60 \implies d = 6 \]

Substitute \( d = 6 \) into equation (1):

\[ 2a + 9(6) = 80 \implies 2a + 54 = 80 \implies 2a = 26 \implies a = 13 \]

Answer: The first term is \( a = 13 \) and the common difference is \( d = 6 \).

(ii) A geometric progression has first term \( a \), common ratio \( r \), and sum to infinity \( 6 \). A second geometric progression has first term \( 2a \), common ratio \( r^2 \), and sum to infinity \( 7 \). Find the values of \( a \) and \( r \).

For a geometric progression, the sum to infinity is given by:

\[ S_\infty = \frac{a}{1-r} \]

For the first progression:

\[ \frac{a}{1-r} = 6 \implies a = 6(1-r) \tag{1} \]

For the second progression:

\[ \frac{2a}{1-r^2} = 7 \implies 2a = 7(1-r^2) \implies a = \frac{7(1-r^2)}{2} \tag{2} \]

Equate equations (1) and (2):

\[ 6(1-r) = \frac{7(1-r^2)}{2} \]

Multiply through by 2:

\[ 12(1-r) = 7(1-r^2) \]

Expand both sides:

\[ 12 - 12r = 7 - 7r^2 \]

Rearrange into standard form:

\[ 7r^2 - 12r - 5 = 0 \]

Solve using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 7, \, b = -12, \, c = -5 \]

\[ r = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(7)(-5)}}{2(7)} = \frac{12 \pm \sqrt{144 + 140}}{14} = \frac{12 \pm \sqrt{284}}{14} \]

\[ r = \frac{12 + \sqrt{284}}{14} \quad \text{or} \quad r = \frac{12 - \sqrt{284}}{14} \]

Substitute \( r \) back into equation (1) to find \( a \).

Question 5

(i) A curve has equation \( y = 3x - \frac{4}{x} \) and passes through the points \( A(1, -1) \) and \( B(4, 11) \). At each of the points \( C \) and \( D \) on the curve, the tangent is parallel to \( AB \). Find the equation of the perpendicular bisector of \( CD \).

1. Gradient of the tangent at any point on the curve:

\[ \frac{dy}{dx} = 3 + \frac{4}{x^2} \]

Gradient of line \( AB \):

\[ m_{AB} = \frac{11 - (-1)}{4 - 1} = \frac{12}{3} = 4 \]

Set the derivative equal to \( 4 \) (tangent is parallel to \( AB \)):

\[ 3 + \frac{4}{x^2} = 4 \implies \frac{4}{x^2} = 1 \implies x^2 = 4 \implies x = \pm 2 \]

For \( x = 2 \):

\[ y = 3(2) - \frac{4}{2} = 6 - 2 = 4 \implies C(2, 4) \]

For \( x = -2 \):

\[ y = 3(-2) - \frac{4}{-2} = -6 + 2 = -4 \implies D(-2, -4) \]

The midpoint of \( CD \):

\[ \text{Midpoint} = \left(\frac{2 + (-2)}{2}, \frac{4 + (-4)}{2}\right) = (0, 0) \]

The gradient of \( CD \):

\[ m_{CD} = \frac{-4 - 4}{-2 - 2} = \frac{-8}{-4} = 2 \]

The perpendicular gradient is:

\[ m = -\frac{1}{2} \]

Equation of the perpendicular bisector:

\[ y - 0 = -\frac{1}{2}(x - 0) \implies y = -\frac{1}{2}x \]

Answer: \( y = -\frac{1}{2}x \).

(ii) A curve passes through the point \( P(5, 1) \) and has equation \( y = 3 - \frac{10}{x} \). Show that the equation of the normal is \( 5x + 2y = 27 \).

Derivative of the curve:

\[ \frac{dy}{dx} = \frac{10}{x^2} \]

Gradient of the tangent at \( P(5, 1) \):

\[ m_{\text{tangent}} = \frac{2}{5} \]

Gradient of the normal:

\[ m_{\text{normal}} = -\frac{5}{2} \]

Equation of the normal:

\[ y - 1 = -\frac{5}{2}(x - 5) \implies 5x + 2y = 27 \]

(iii) Find the point \( Q \).

Solving the quadratic equation:

\[ 5x^2 - 21x - 20 = 0 \implies x = 5 \, \text{or} \, x = -\frac{4}{5} \]

At \( x = -\frac{4}{5} \):

\[ y = 15.5 \implies Q = \left(-\frac{4}{5}, 15.5\right) \]

(iii) For the watermelon problem:

Value of \( k = 0.0032 \), rate of mass increase = \( 0.096 \, \text{kg/day} \).

Review:

Question 6(i)

The diagram shows the curve \( y = 6x - x^2 \) and the line \( y = 5 \). Find the area of the shaded region.

Solution:

Find the points of intersection between \( y = 6x - x^2 \) and \( y = 5 \):

\[ 6x - x^2 = 5 \]

Rearrange into standard quadratic form:

\[ x^2 - 6x + 5 = 0 \]

Factorize:

\[ (x - 5)(x - 1) = 0 \implies x = 1 \quad \text{and} \quad x = 5 \]

The area of the shaded region is the integral of the difference between the curve \( y = 6x - x^2 \) and the line \( y = 5 \) from \( x = 1 \) to \( x = 5 \):

\[ \text{Area} = \int_{1}^{5} \left((6x - x^2) - 5\right) \, dx \]

Simplify the integrand:

\[ (6x - x^2) - 5 = -x^2 + 6x - 5 \]

Integrate:

\[ \int (-x^2 + 6x - 5) \, dx = -\frac{x^3}{3} + 3x^2 - 5x \]

Evaluate the definite integral:

\[ \left[-\frac{x^3}{3} + 3x^2 - 5x \right]_1^5 \]

At \( x = 5 \):

\[ -\frac{5^3}{3} + 3(5^2) - 5(5) = -\frac{125}{3} + 75 - 25 = \frac{-125 + 225 - 75}{3} = \frac{25}{3} \]

At \( x = 1 \):

\[ -\frac{1^3}{3} + 3(1^2) - 5(1) = -\frac{1}{3} + 3 - 5 = -\frac{1}{3} - 2 = -\frac{7}{3} \]

Subtract:

\[ \text{Area} = \frac{25}{3} - \left(-\frac{7}{3}\right) = \frac{25}{3} + \frac{7}{3} = \frac{32}{3} \]

Answer: The area of the shaded region is \( \frac{32}{3} \) square units.

Question 6(ii)

The diagram shows parts of the curves \( y = (4x + 1)^{\frac{1}{2}} \) and \( y = \frac{1}{2}x^2 + 1 \), intersecting at points \( P(0, 1) \) and \( Q(2, 3) \).

(i) Find \( \alpha \), the angle between the tangents to the curves at \( Q \).

Find the derivatives of both curves:

- For \( y = (4x + 1)^{\frac{1}{2}} \):

\[ \frac{dy}{dx} = \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 = \frac{2}{\sqrt{4x + 1}} \]

At \( Q(2, 3) \):

\[ \frac{dy}{dx} = \frac{2}{\sqrt{4(2) + 1}} = \frac{2}{\sqrt{9}} = \frac{2}{3} \]

- For \( y = \frac{1}{2}x^2 + 1 \):

\[ \frac{dy}{dx} = x \]

At \( Q(2, 3) \):

\[ \frac{dy}{dx} = 2 \]

The gradients of the tangents are \( m_1 = \frac{2}{3} \) and \( m_2 = 2 \). The angle \( \alpha \) between the tangents is given by:

\[ \tan \alpha = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]

Substitute values:

\[ \tan \alpha = \left| \frac{2 - \frac{2}{3}}{1 + \frac{2}{3} \cdot 2} \right| = \left| \frac{\frac{6}{3} - \frac{2}{3}}{1 + \frac{4}{3}} \right| = \left| \frac{\frac{4}{3}}{\frac{7}{3}} \right| = \frac{4}{7} \]

Find \( \alpha \):

\[ \alpha = \arctan \left(\frac{4}{7}\right) \]

Using a calculator:

\[ \alpha \approx 29.74^\circ \]

Answer: \( \alpha = 29.7^\circ \) (to 3 significant figures).

(ii) Find by integration the area of the shaded region.

The shaded region is the area between the curves \( y = (4x + 1)^{\frac{1}{2}} \) and \( y = \frac{1}{2}x^2 + 1 \) from \( x = 0 \) to \( x = 2 \):

\[ \text{Area} = \int_0^2 \left((4x + 1)^{\frac{1}{2}} - \left(\frac{1}{2}x^2 + 1\right)\right) dx \]

Integrate each term separately:

\[ \int_0^2 (4x + 1)^{\frac{1}{2}} dx - \int_0^2 \left(\frac{1}{2}x^2 + 1\right) dx \]

- For \( \int_0^2 (4x + 1)^{\frac{1}{2}} \, dx \):

\[ \frac{1}{6} \left(27 - 1\right) = \frac{13}{3} \]

- For \( \int_0^2 \left(\frac{1}{2}x^2 + 1\right) dx \):

\[ \frac{10}{3} \]

Subtract:

\[ \text{Area} = \frac{13}{3} - \frac{10}{3} = 1 \]

Answer: The area of the shaded region is \( 1 \, \text{square unit.} \)

Application Problems

The area under a curve is found by calculating the definite integral of the function between two limits \(a\) and \(b\). The general formula is:

\[ \text{Area} = \int_{a}^{b} f(x) \, dx \]

Where \( f(x) \) is the function and \(a\) and \(b\) are the limits of integration.

Let's look at a few examples of how to compute the area under a curve:

1. Example 1: Find the area under the curve of \(y = x^2\) from \(x = 0\) to \(x = 2\).

The integral we need to evaluate is:

\[ \text{Area} = \int_{0}^{2} x^2 \, dx \]

Step-by-step solution:

1. Set up the integral:

\[ \text{Area} = \int_{0}^{2} x^2 \, dx \]

2. Integrate:

\[ \int x^2 \, dx = \frac{x^3}{3} \]

3. Apply the limits:

\[ \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3} \]

The area under the curve is \( \frac{8}{3} \) sqUnits.

2. Example 2: Find the area under the curve of \( y = 3x^2 + 2x \) from \(x = 1\) to \(x = 3\).

The integral we need to evaluate is:

\[ \text{Area} = \int_{1}^{3} \left( 3x^2 + 2x \right) \, dx \]

Step-by-step solution:

1. Set up the integral:

\[ \text{Area} = \int_{1}^{3} \left( 3x^2 + 2x \right) \, dx \]

2. Integrate each term:

\[ \int 3x^2 \, dx = x^3, \quad \int 2x \, dx = x^2 \]

3. Apply the limits:

\[ \left[ x^3 \right]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26 \] \[ \left[ x^2 \right]_{1}^{3} = 3^2 - 1^2 = 9 - 1 = 8 \]

4. Final result:

\[ \text{Area} = 26 + 8 = 34 sqUnits \]

Solution:

Volume Integrals - A-Level Cambridge

Introduction to Volume Integrals

In mathematics, particularly in calculus, volume integrals are used to find the volume of solids that are defined by a function or a surface. These integrals are typically solved using the method of **integration** and **geometry**. Volume integrals are commonly encountered when determining the volume of regions in **3D space** using techniques such as the **disk method**, **washer method**, and **cylindrical shells**.

General Formulae:

The volume \( V \) of a solid of revolution about the x-axis can be calculated using the following formula:

\[ V = \pi \int_{a}^{b} \left( f(x) \right)^2 \, dx \] where \( f(x) \) is the function defining the curve and \( a \) and \( b \) are the bounds of integration along the x-axis.

The volume of a solid of revolution about the y-axis can be calculated with the formula:

\[ V = \pi \int_{c}^{d} \left( g(y) \right)^2 \, dy \] where \( g(y) \) is the function describing the curve, and \( c \) and \( d \) are the bounds along the y-axis.

Disk Method:

When the solid is generated by rotating a function \( y = f(x) \) about the x-axis (or a function \( x = g(y) \) about the y-axis), the volume is given by the formula:

\[ V = \pi \int_{a}^{b} \left( f(x) \right)^2 \, dx \] The function \( f(x) \) represents the radius of the disk at each value of \( x \).

Washer Method:

When there is a hole in the middle of the solid, we use the washer method, which involves subtracting the inner radius from the outer radius:

\[ V = \pi \int_{a}^{b} \left( \left( f(x) \right)^2 - \left( g(x) \right)^2 \right) dx \] where \( f(x) \) is the outer radius and \( g(x) \) is the inner radius of the washer at each point along the axis.

Cylindrical Shells Method:

For solids generated by rotating a function \( y = f(x) \) about the y-axis, the volume can also be found using the cylindrical shells method:

\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where \( x \) represents the radius and \( f(x) \) represents the height of each cylindrical shell.

Example 1: Using the Disk Method

Find the volume of the solid formed by rotating the region bounded by \( y = x^2 \) and the x-axis from \( x = 0 \) to \( x = 1 \) about the x-axis.

Solution:

The formula for volume is:

\[ V = \pi \int_{a}^{b} \left( f(x) \right)^2 \, dx \] where \( f(x) = x^2 \) and the bounds are from \( x = 0 \) to \( x = 1 \). Thus, we have:

\[ V = \pi \int_{0}^{1} \left( x^2 \right)^2 \, dx = \pi \int_{0}^{1} x^4 \, dx \] Integrating \( x^4 \), we get: \[ V = \pi \left[ \frac{x^5}{5} \right]_{0}^{1} = \pi \left( \frac{1}{5} - 0 \right) = \frac{\pi}{5} \] So, the volume is \( \frac{\pi}{5} \) cubic units.

Example 2: Using the Washer Method

Find the volume of the solid formed by rotating the region bounded by \( y = x^2 \) and \( y = 1 \) about the x-axis from \( x = 0 \) to \( x = 1 \).

Solution:

Here, the volume is found using the washer method. The outer radius is \( R(x) = 1 \) and the inner radius is \( r(x) = x^2 \). The volume formula is:

\[ V = \pi \int_{0}^{1} \left( R(x)^2 - r(x)^2 \right) \, dx \] Substituting the functions \( R(x) = 1 \) and \( r(x) = x^2 \), we get: \[ V = \pi \int_{0}^{1} \left( 1^2 - \left( x^2 \right)^2 \right) \, dx = \pi \int_{0}^{1} \left( 1 - x^4 \right) \, dx \] Now, integrate: \[ V = \pi \left[ x - \frac{x^5}{5} \right]_{0}^{1} = \pi \left( 1 - \frac{1}{5} \right) = \pi \times \frac{4}{5} \] So, the volume is \( \frac{4\pi}{5} \) cubic units.

Example 3: Using the Cylindrical Shells Method

Find the volume of the solid formed by rotating the region bounded by \( y = x^2 \) and the y-axis from \( x = 0 \) to \( x = 1 \) about the y-axis.

Solution:

The formula for volume using cylindrical shells is:

\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] Here, \( f(x) = x^2 \) and the bounds are from \( x = 0 \) to \( x = 1 \). So, we have:

\[ V = 2\pi \int_{0}^{1} x \cdot x^2 \, dx = 2\pi \int_{0}^{1} x^3 \, dx \] Now, integrate \( x^3 \): \[ V = 2\pi \left[ \frac{x^4}{4} \right]_{0}^{1} = 2\pi \left( \frac{1}{4} - 0 \right) = \frac{\pi}{2} \] So, the volume is \( \frac{\pi}{2} \) cubic units.

Practice Exercises

Try solving the following volume problems:

1. Find the volume of the solid formed by rotating the region bounded by \( y = 2x \) and the x-axis from \( x = 0 \) to \( x = 3 \) about the x-axis using the disk method.
2. Find the volume of the solid formed by rotating the region bounded by \( y = x^2 \) and \( y = 4 \) about the x-axis from \( x = 0 \) to \( x = 2 \) using the washer method.
3. Find the volume of the solid formed by rotating the region bounded by \( y = \sqrt{x} \) and the x-axis from \( x = 0 \) to \( x = 4 \) about the y-axis using the cylindrical shells method.
4. Find the volume of the solid formed by rotating the region bounded by \( y = x^3 \) and the x-axis from \( x = 0 \) to \( x = 1 \) about the x-axis using the disk method.

Answer Key (Optional):

1. Volume = \( 54\pi \) cubic units
2. Volume = \( \frac{28\pi}{5} \) cubic units
3. Volume = \( 32\pi \) cubic units
4. Volume = \( \frac{\pi}{5} \) cubic units

Worksheet-Integrals

Evaluate the following integrals:

1. a) \[ \int_{0}^{1} \left( 2x^2 - 3x + 5 \right) \, dx \]
2. b) \[ \int_{1}^{2} \left( 4x^3 - 2x + 1 \right) \, dx \]
3. c) \[ \int_{0}^{3} \left( x^2 + 2x - 1 \right) \, dx \]
4. d) \[ \int_{0}^{2} \left( x^2 - 4x + 4 \right) \, dx \]
5. e) \[ \int_{1}^{3} \left( 2x^3 - 5x^2 + 4x - 1 \right) \, dx \]
6. f) \[ \int_{2}^{4} \left( 3x^2 + 2x - 7 \right) \, dx \]
7. g) \[ \int_{0}^{2} \left( x^4 - x^3 + 3x^2 - 5x \right) \, dx \]
8. h) \[ \int_{-1}^{1} \left( x^3 - x \right) \, dx \]
9. i) \[ \int_{0}^{\pi} \left( \sin(x) + \cos(x) \right) \, dx \]
10. j) \[ \int_{0}^{\pi/2} \left( \tan(x) \right) \, dx \]
11. k) \[ \int_{0}^{1} \left( \frac{1}{x} \right) \, dx \]
12. l) \[ \int_{1}^{3} \left( \frac{4}{x^2} \right) \, dx \]
13. m) \[ \int_{1}^{4} \left( \frac{1}{x^3} \right) \, dx \]
14. n) \[ \int_{2}^{5} \left( x^2 + 2x + 1 \right) \, dx \]
15. o) \[ \int_{0}^{\infty} \left( e^{-x} \right) \, dx \]
16. p) \[ \int_{-2}^{2} \left( x^2 - 4x \right) \, dx \]
17. q) \[ \int_{0}^{1} \left( x^3 + 2x^2 - x + 4 \right) \, dx \]
18. r) \[ \int_{1}^{2} \left( \frac{1}{x^2} - 3 \right) \, dx \]
19. s) \[ \int_{0}^{2} \left( 4x^2 - 6x + 2 \right) \, dx \]
20. t) \[ \int_{-3}^{0} \left( 3x^2 - 5x + 4 \right) \, dx \]

Answer Key:

1. a) \( \frac{19}{6} \)
2. b) \( 13 \)
3. c) \( 15 \)
4. d) \( \frac{8}{3} \)
5. e) \( \frac{77}{2} \)
6. f) \( 54 \)
7. g) \( \frac{32}{5} \)
8. h) \( 0 \)
9. i) \( 2 \)
10. j) \( \ln(1 + \sqrt{2}) \)
11. k) \( \infty \) (Improper Integral)
12. l) \( -\frac{1}{3} \)
13. m) \( \frac{1}{2} \)
14. n) \( 27 \)
15. o) \( 1 \)
16. p) \( \frac{16}{3} \)
17. q) \( 7 \)
18. r) \( \frac{1}{2} \)
19. s) \( \frac{8}{3} \)
20. t) \( 66 \)

Solution:

The curve is given by the equation:

\[ y = 4 - 2\sqrt{x} \]

Part (a): Find the equation of the normal \( PQ \)

Step 1: Differentiate \( y \) to find the gradient of the tangent:

\[ \frac{dy}{dx} = -\frac{2}{2\sqrt{x}} = -\frac{1}{\sqrt{x}} \]

At \( P(16, -4) \):

\[ m_{\text{tangent}} = -\frac{1}{\sqrt{16}} = -\frac{1}{4} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{-\frac{1}{4}} = 4 \]

The equation of the normal is:

\[ y - (-4) = 4(x - 16) \]

Simplify:

\[ y + 4 = 4x - 64 \implies y = 4x - 68 \]

Part (b): Find the coordinates of \( Q \)

At \( Q \), the normal meets the x-axis (\( y = 0 \)). Substituting \( y = 0 \):

\[ 0 = 4x - 68 \implies x = 17 \]

So, the coordinates of \( Q \) are:

\[ Q(17, 0) \]

Solution:

The curve is given by the equation:

\[ y = 2x - \frac{10}{x^2} + 8 \]

Part (a): Find \(\frac{dy}{dx}\)

Differentiate term by term:

\[ \frac{dy}{dx} = 2 - \frac{d}{dx}\left(\frac{10}{x^2}\right) \]

Using the power rule:

\[ \frac{dy}{dx} = 2 + 20x^{-3} = 2 + \frac{20}{x^3} \]

Part (b): Show that the normal passes through \( (0, -3) \)

At \( \left(-4, -\frac{5}{8}\right) \):

\[ m_{\text{tangent}} = 2 + \frac{20}{(-4)^3} = 2 + \frac{20}{-64} = 2 - \frac{5}{16} = \frac{32}{16} - \frac{5}{16} = \frac{27}{16} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{\frac{27}{16}} = -\frac{16}{27} \]

The equation of the normal is:

\[ y - \left(-\frac{5}{8}\right) = -\frac{16}{27}(x - (-4)) \]

Simplify:

\[ y + \frac{5}{8} = -\frac{16}{27}(x + 4) \]

Substitute \( x = 0 \):

\[ y + \frac{5}{8} = -\frac{16}{27}(4) = -\frac{64}{27} \]

Simplify further:

\[ y = -\frac{64}{27} - \frac{5}{8} \]

Converting to a common denominator:

\[ y = -3 \implies \text{Passes through } (0, -3) \]

Solution:

Differentiate the given equation w.r.t x:

\[ y = \frac{3x^5 - 7}{4x} \]

Step 1: Rewrite the equation:

\[ y = \frac{3x^5}{4x} - \frac{7}{4x} = \frac{3x^4}{4} - \frac{7}{4x} \]

Step 2: Differentiate term by term:

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{3x^4}{4}\right) - \frac{d}{dx}\left(\frac{7}{4x}\right) \]

\[ \frac{dy}{dx} = \frac{3}{4} \cdot 4x^3 - \frac{7}{4} \cdot \frac{-1}{x^2} \]

\[ \frac{dy}{dx} = 3x^3 + \frac{7}{4x^2} \]

Final Answer:

\[ \frac{dy}{dx} = 3x^3 + \frac{7}{4x^2} \]

Solution:

Find the Gradient of the curve, at x = 2:

\[ y = \frac{8}{4x - 5} \]

Step 1: Differentiate using the chain rule:

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{8}{u}\right) \cdot \frac{du}{dx}, \quad \text{where } u = 4x - 5 \]

\[ \frac{dy}{dx} = \frac{-8}{(4x - 5)^2} \cdot \frac{d}{dx}(4x - 5) \]

\[ \frac{dy}{dx} = \frac{-8}{(4x - 5)^2} \cdot 4 = \frac{-32}{(4x - 5)^2} \]

Step 2: Find the gradient at \( x = 2 \):

Substitute \( x = 2 \):

\[ \frac{dy}{dx} = \frac{-32}{(4(2) - 5)^2} = \frac{-32}{(8 - 5)^2} = \frac{-32}{9} \]

Final Answer:

The gradient at \( x = 2 \) is:

\[ \frac{-32}{9} \]

Solution:

\[ y = 3x^3 - 3x^2 + x - 7 \]

Part (a): Show that the gradient of the curve is never negative.

Step 1: Differentiate \( y \):

\[ \frac{dy}{dx} = 9x^2 - 6x + 1 \]

Step 2: Analyze the quadratic expression \( 9x^2 - 6x + 1 \):

The discriminant of a quadratic equation \( ax^2 + bx + c \) is given by:

\[ \Delta = b^2 - 4ac \]

Here, \( a = 9 \), \( b = -6 \), and \( c = 1 \):

\[ \Delta = (-6)^2 - 4(9)(1) = 36 - 36 = 0 \]

Since the discriminant is \( 0 \), the quadratic has a single root, and the parabola opens upwards (as \( a > 0 \)).

Step 3: Verify that \( \frac{dy}{dx} \geq 0 \):

The vertex of the parabola occurs at:

\[ x = -\frac{b}{2a} = -\frac{-6}{2(9)} = \frac{1}{3} \]

At the vertex, \( \frac{dy}{dx} = 0 \). For all other values of \( x \), \( \frac{dy}{dx} > 0 \) because the parabola opens upwards.

Final Answer:

The gradient of the curve is never negative.

Reference-https://mathz.org/Alevels/week12/week12.html

Differentiation Practice Worksheet

Below are 15 questions on differentiation. Solve step by step and show your working:

Part A: Basic Differentiation

Part B: Applications of Differentiation

Part C: Advanced Differentiation

Part D: Mixed Problems

End of Chapter Worsheet:

Binomial Expansion