Comprehensive Analysis of SS3 Lesson Notes for Second Term: A Guide to Key Topics in Mathematics

Mathematics at the SS3 level can seem intimidating, but with the right breakdown, any student can grasp these concepts and apply them efficiently in exams and real-life situations. In this comprehensive analysis, we will cover essential topics that students encounter in their second-term mathematics curriculum. These topics span a range of areas including finance, coordinate geometry, differentiation, and integration, all of which are foundational to higher-level math and real-world applications.

Review of First Term Work

Before diving into the second-term topics, it’s essential to first understand the foundation laid in the previous term. The review of first-term work provides students with a chance to revisit key areas, solidify their understanding, and make connections to the more advanced concepts covered in the current term.

Bonds and Debentures
Bonds and debentures are both debt instruments used by companies and governments to raise funds. A bond is essentially a loan made by the investor to the issuer, which is repaid with interest over time. Debentures, on the other hand, are a type of unsecured bond, meaning they are not backed by physical assets but rather by the general creditworthiness of the issuer.
Example: If a company issues a bond worth ₦10,000 with a 5% interest rate, the investor will receive ₦500 annually until the bond matures.

Shares
Shares represent ownership in a company. When you purchase shares, you become a shareholder and gain certain rights, such as voting in company decisions and receiving dividends.
Example: If you buy 100 shares in a company priced at ₦50 each, you own part of the company and are entitled to a portion of the company’s profits, usually in the form of dividends.

Rates
Rates refer to the percentage of interest or fees charged for a service, typically in financial contexts. This can include interest rates on loans or savings accounts.
Example: A savings account offering a 3% annual interest rate means you earn 3% of the amount you have in the account annually.

Income Tax
Income tax is a tax imposed on an individual’s earnings. The amount paid depends on the individual’s income level and tax bracket.
Example: If you earn ₦500,000 annually and your tax rate is 10%, you will pay ₦50,000 in income tax.

Value Added Tax (VAT)
VAT is a consumption tax placed on goods and services. The rate varies by country, but the consumer ultimately bears the cost.
Example: If an item costs ₦1,000 and VAT is 10%, the total price will be ₦1,100.

Week 2: Coordinate Geometry – The Straight Line

Topic 1: The General Equation of a Straight Line

In coordinate geometry, the equation of a straight line can be written in several forms. The most common forms are:

1. Slope-Intercept Form:
   $y=mx+c$ Where:
   • $m$ is the slope of the line (the rate of change of y with respect to x),
   • $c$ is the y-intercept (the point where the line crosses the y-axis).

   Example: If the equation of the line is $y=2x+3$, it means:

   • The slope $m=2$, indicating that for every unit increase in x, y increases by 2.
   • The line intersects the y-axis at $y=3$.
2. Point-Slope Form:
   y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​) Where:
   • (x1,y1)(x_1, y_1)(x1​,y1​) is a point on the line,
   • $m$ is the slope.

   Example: If the line passes through the point $(3,4)$ and has a slope of $2$, the equation is: $y-4=2(x-3)$

3. General Form of a Line:
   $Ax+By+C=0$ Where $A$, $B$, and $C$ are constants.Example: An equation like $2x-3y+4=0$ is a straight line equation in general form.

Topic 2: Slope of a Line

The slope of a line is a measure of how steep the line is. It is calculated using two points on the line. The formula for calculating the slope $m$ of a line that passes through two points (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​) is:

m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2​−x1​y2​−y1​​

This formula tells you how much $y$ increases or decreases as $x$ changes between the two points.

Example:
For the points $(1,2)$ and $(4,8)$, the slope of the line is:

m=8−24−1=63=2m = \frac{8 – 2}{4 – 1} = \frac{6}{3} = 2m=4−18−2​=36​=2

Thus, the slope of the line is 2.

Topic 3: Equation of a Line Given Two Points

If you are given two points, you can find the equation of the straight line passing through them. First, calculate the slope $m$ using the formula above. Then, use the point-slope form of the equation of the line.

Steps:

1. Find the slope $m$ of the line using the formula.
2. Use one of the points and the slope in the point-slope form to find the equation of the line.

Example:
Given the points $(1,2)$ and $(4,8)$, the slope is 2. Using the point $(1,2)$ in the point-slope form:

$$y-2=2(x-1)$$

Simplifying:

$$y-2=2x-2\Rightarrow y=2x$$

Thus, the equation of the line is $y=2x$.

Topic 4: Parallel and Perpendicular Lines

In coordinate geometry, understanding the relationship between parallel and perpendicular lines is important.

• Parallel Lines: Two lines are parallel if they have the same slope. That is, if m1=m2m_1 = m_2m1​=m2​, then the lines are parallel.
• Perpendicular Lines: Two lines are perpendicular if the product of their slopes is -1. That is, if m1×m2=−1m_1 \times m_2 = -1m1​×m2​=−1, the lines are perpendicular.

Example: For lines with slopes m1=3m_1 = 3m1​=3 and m2=−13m_2 = -\frac{1}{3}m2​=−31​, the lines are perpendicular because:

3×−13=−13 \times -\frac{1}{3} = -13×−31​=−1

Topic 5: Distance Between Two Points

To find the distance $d$ between two points (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​) on a coordinate plane, use the distance formula:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​

Example:
For the points $(1,2)$ and $(4,6)$, the distance is:

d=(4−1)2+(6−2)2=9+16=25=5d = \sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5d=(4−1)2+(6−2)2​=9+16​=25​=5

Week 3: Further Exploration of Coordinate Geometry

Topic 1: The Equation of a Circle

The equation of a circle in the coordinate plane is fundamental in coordinate geometry. It helps us represent circular shapes and understand their properties.

Equation of a Circle

The general equation of a circle with center $(h,k)$ and radius $r$ is:

(x−h)2+(y−k)2=r2(x – h)^2 + (y – k)^2 = r^2(x−h)2+(y−k)2=r2

Where:

• $(h,k)$ are the coordinates of the center of the circle,
• $r$ is the radius, the distance from the center to any point on the circle.

Example:
Consider a circle with center $(3,4)$ and radius 5. The equation of this circle is:

(x−3)2+(y−4)2=25(x – 3)^2 + (y – 4)^2 = 25(x−3)2+(y−4)2=25

This equation tells us that every point $(x,y)$ that satisfies the equation lies on the circle.

Important Concepts Related to Circles

• Radius: The distance from the center of the circle to any point on the circle.
• Diameter: The longest distance across the circle, passing through the center. The diameter is twice the radius.
• Circumference: The total distance around the circle, calculated as $C=2\pi r$.

Topic 2: Equation of a Tangent to a Circle

A tangent to a circle is a straight line that touches the circle at exactly one point. The point where the tangent touches the circle is called the point of tangency.

Equation of a Tangent

If we have a circle with the equation (x−h)2+(y−k)2=r2(x – h)^2 + (y – k)^2 = r^2(x−h)2+(y−k)2=r2, and we are given a point P(x1,y1)P(x_1, y_1)P(x1​,y1​) on the circle, the equation of the tangent at that point is:

(x1−h)(x−h)+(y1−k)(y−k)=r2(x_1 – h)(x – h) + (y_1 – k)(y – k) = r^2(x1​−h)(x−h)+(y1​−k)(y−k)=r2

This equation is derived from the fact that the radius to the point of tangency is perpendicular to the tangent.

Example:
Consider a circle with the equation (x−2)2+(y−3)2=16(x – 2)^2 + (y – 3)^2 = 16(x−2)2+(y−3)2=16 and a point $P(6,3)$ on the circle. The equation of the tangent at point $P(6,3)$ is:

$$(6-2)(x-2)+(3-3)(y-3)=16$$

Simplifying:

$$4(x-2)=16\Rightarrow x=6$$

Thus, the equation of the tangent line is $x=6$, a vertical line passing through the point $P(6,3)$.

Topic 3: Application of Coordinate Geometry in Geometry Problems

Coordinate geometry is widely used to solve complex geometry problems. Some common applications include finding the area of polygons, determining the length of line segments, and analyzing geometric shapes such as triangles, quadrilaterals, and circles.

Example 1: Area of a Triangle

The area of a triangle with vertices (x1,y1)(x_1, y_1)(x1​,y1​), (x2,y2)(x_2, y_2)(x2​,y2​), and (x3,y3)(x_3, y_3)(x3​,y3​) in the coordinate plane can be calculated using the formula:

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|Area=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

Example:
Given the vertices $A(1,2)$, $B(4,5)$, and $C(7,2)$, the area of the triangle is:

Area=12∣1(5−2)+4(2−2)+7(2−5)∣=12∣3+0−21∣=12×18=9\text{Area} = \frac{1}{2} \left| 1(5 – 2) + 4(2 – 2) + 7(2 – 5) \right| = \frac{1}{2} \left| 3 + 0 – 21 \right| = \frac{1}{2} \times 18 = 9Area=21​∣1(5−2)+4(2−2)+7(2−5)∣=21​∣3+0−21∣=21​×18=9

Thus, the area of the triangle is 9 square units.

Example 2: Distance Between Two Points

The distance between two points (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​) in the coordinate plane is given by the formula:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​

Example:
Given the points $A(1,2)$ and $B(4,6)$, the distance between the points is:

d=(4−1)2+(6−2)2=9+16=25=5d = \sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5d=(4−1)2+(6−2)2​=9+16​=25​=5

Thus, the distance between the points $A$ and $B$ is 5 units.

Topic 4: The Use of Slope in Real-Life Applications

In real-life situations, coordinate geometry, particularly the concept of slope, can be used to solve problems involving rates of change. For example, in economics, the slope of a line might represent the rate of profit increase over time, or in physics, the slope of a line might represent speed or velocity.

Differentiating Algebraic Functions

1. Differentiating Polynomials

A polynomial function is an algebraic expression consisting of terms with non-negative integer exponents. The derivative of each term in a polynomial is found by applying the power rule.

Example 1:

Differentiate the function f(x)=4×3−2×2+5x−7f(x) = 4x^3 – 2x^2 + 5x – 7f(x)=4x3−2x2+5x−7.

Solution:
We differentiate each term using the power rule.

f′(x)=ddx(4×3)−ddx(2×2)+ddx(5x)−ddx(7)f'(x) = \frac{d}{dx} (4x^3) – \frac{d}{dx} (2x^2) + \frac{d}{dx} (5x) – \frac{d}{dx} (7)f′(x)=dxd​(4x3)−dxd​(2x2)+dxd​(5x)−dxd​(7) f′(x)=4⋅3×2−2⋅2x+5−0f'(x) = 4 \cdot 3x^2 – 2 \cdot 2x + 5 – 0f′(x)=4⋅3x2−2⋅2x+5−0 f′(x)=12×2−4x+5f'(x) = 12x^2 – 4x + 5f′(x)=12x2−4x+5

Thus, the derivative of f(x)=4×3−2×2+5x−7f(x) = 4x^3 – 2x^2 + 5x – 7f(x)=4x3−2x2+5x−7 is f′(x)=12×2−4x+5f'(x) = 12x^2 – 4x + 5f′(x)=12x2−4x+5.

2. Differentiating Rational Functions

Rational functions are functions that involve a ratio of two polynomials. To differentiate rational functions, we apply the quotient rule.

Example 2:

Differentiate f(x)=3×2+5×3+2xf(x) = \frac{3x^2 + 5}{x^3 + 2x}f(x)=x3+2x3x2+5​.

Solution:
Let g(x)=3×2+5g(x) = 3x^2 + 5g(x)=3x2+5 and h(x)=x3+2xh(x) = x^3 + 2xh(x)=x3+2x. The quotient rule tells us that:

f′(x)=g′(x)⋅h(x)−g(x)⋅h′(x)(h(x))2f'(x) = \frac{g'(x) \cdot h(x) – g(x) \cdot h'(x)}{(h(x))^2}f′(x)=(h(x))2g′(x)⋅h(x)−g(x)⋅h′(x)​

First, differentiate $g(x)$ and $h(x)$:

g′(x)=6xg'(x) = 6xg′(x)=6x h′(x)=3×2+2h'(x) = 3x^2 + 2h′(x)=3x2+2

Now, apply the quotient rule:

f′(x)=(6x)⋅(x3+2x)−(3×2+5)⋅(3×2+2)(x3+2x)2f'(x) = \frac{(6x) \cdot (x^3 + 2x) – (3x^2 + 5) \cdot (3x^2 + 2)}{(x^3 + 2x)^2}f′(x)=(x3+2x)2(6x)⋅(x3+2x)−(3x2+5)⋅(3x2+2)​

Simplify the expression:

f′(x)=6x(x3+2x)−(3×2+5)(3×2+2)(x3+2x)2f'(x) = \frac{6x(x^3 + 2x) – (3x^2 + 5)(3x^2 + 2)}{(x^3 + 2x)^2}f′(x)=(x3+2x)26x(x3+2x)−(3x2+5)(3x2+2)​

You can further expand and simplify this expression, but the key idea is to use the quotient rule for rational functions.

3. Differentiating Higher Powers of $x$

When differentiating higher powers of $x$, we still apply the power rule. If the exponent is negative or a fraction, the power rule still holds.

Example 3:

Differentiate f(x)=x−2+1xf(x) = x^{-2} + \frac{1}{x}f(x)=x−2+x1​.

Solution:
Apply the power rule to each term:

f′(x)=ddx(x−2)+ddx(1x)f'(x) = \frac{d}{dx}(x^{-2}) + \frac{d}{dx}\left(\frac{1}{x}\right)f′(x)=dxd​(x−2)+dxd​(x1​) f′(x)=−2x−3−x−2f'(x) = -2x^{-3} – x^{-2}f′(x)=−2x−3−x−2

Thus, the derivative of f(x)=x−2+1xf(x) = x^{-2} + \frac{1}{x}f(x)=x−2+x1​ is f′(x)=−2x−3−x−2f'(x) = -2x^{-3} – x^{-2}f′(x)=−2x−3−x−2.

Practical Applications of Differentiation

Differentiation is used in many real-world applications, including:

• Physics: To find the velocity of an object by differentiating its position function.
• Economics: To find the marginal cost or marginal revenue by differentiating the cost and revenue functions.
• Engineering: In determining the rate of change of a system or process.

Differentiation provides insights into how quantities change over time, and understanding how to differentiate algebraic functions is key to solving many real-life problems.

1. The Chain Rule

The chain rule is used to differentiate composite functions—functions that are made up of two or more other functions. If you have a function $y=f(g(x))$, the chain rule allows you to differentiate it by multiplying the derivative of the outer function by the derivative of the inner function.

Chain Rule Formula:

If $y=f(g(x))$, then:

dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)dxdy​=f′(g(x))⋅g′(x)

This rule is useful when you have functions nested within each other, like (3×2+2x)5(3x^2 + 2x)^5(3x2+2x)5.

Example 1:

Differentiate y=(2×3+5x)4y = (2x^3 + 5x)^4y=(2x3+5x)4.

Solution:
Let g(x)=2×3+5xg(x) = 2x^3 + 5xg(x)=2x3+5x and f(u)=u4f(u) = u^4f(u)=u4, where $u=g(x)$. Then, the chain rule gives:

dydx=4(2×3+5x)3⋅ddx(2×3+5x)\frac{dy}{dx} = 4(2x^3 + 5x)^3 \cdot \frac{d}{dx}(2x^3 + 5x)dxdy​=4(2x3+5x)3⋅dxd​(2x3+5x)

First, differentiate g(x)=2×3+5xg(x) = 2x^3 + 5xg(x)=2x3+5x:

g′(x)=6×2+5g'(x) = 6x^2 + 5g′(x)=6x2+5

Now, apply the chain rule:

dydx=4(2×3+5x)3⋅(6×2+5)\frac{dy}{dx} = 4(2x^3 + 5x)^3 \cdot (6x^2 + 5)dxdy​=4(2x3+5x)3⋅(6x2+5)

Thus, the derivative is:

dydx=4(2×3+5x)3(6×2+5)\frac{dy}{dx} = 4(2x^3 + 5x)^3(6x^2 + 5)dxdy​=4(2x3+5x)3(6x2+5)

2. Implicit Differentiation

Implicit differentiation is used when the equation is not explicitly solved for one variable in terms of the other (e.g., $x$ and $y$ are mixed). Instead of solving for $y$, you differentiate both sides of the equation with respect to $x$, treating $y$ as an implicit function of $x$.

Steps for Implicit Differentiation:

1. Differentiate both sides of the equation with respect to $x$.
2. When differentiating terms involving $y$, treat $y$ as $y(x)$, so apply the chain rule.
3. Solve for dydx\frac{dy}{dx}dxdy​ after differentiating.

Example 2:

Differentiate the equation x2+y2=25x^2 + y^2 = 25x2+y2=25.

Solution:
Differentiate both sides with respect to $x$:

ddx(x2)+ddx(y2)=ddx(25)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)dxd​(x2)+dxd​(y2)=dxd​(25) 2x+2ydydx=02x + 2y \frac{dy}{dx} = 02x+2ydxdy​=0

Now, solve for dydx\frac{dy}{dx}dxdy​:

dydx=−xy\frac{dy}{dx} = -\frac{x}{y}dxdy​=−yx​

Thus, the derivative of the implicit equation x2+y2=25x^2 + y^2 = 25x2+y2=25 is dydx=−xy\frac{dy}{dx} = -\frac{x}{y}dxdy​=−yx​.

3. Higher-Order Derivatives

The higher-order derivatives refer to the derivatives of derivatives. The first derivative tells you the rate of change, the second derivative tells you the rate of change of the rate of change (i.e., the concavity of the function), and higher derivatives provide more detailed information about the function’s behavior.

Second Derivative:

The second derivative is the derivative of the first derivative. It is denoted as f′′(x)f”(x)f′′(x) or d2dx2[f(x)]\frac{d^2}{dx^2} [f(x)]dx2d2​[f(x)].

Example 3:

Find the second derivative of f(x)=3×4−5×2+7x−1f(x) = 3x^4 – 5x^2 + 7x – 1f(x)=3x4−5x2+7x−1.

Solution:

1. First, find the first derivative:

f′(x)=12×3−10x+7f'(x) = 12x^3 – 10x + 7f′(x)=12x3−10x+7

2. Then, find the second derivative:

f′′(x)=36×2−10f”(x) = 36x^2 – 10f′′(x)=36x2−10

Thus, the second derivative of f(x)=3×4−5×2+7x−1f(x) = 3x^4 – 5x^2 + 7x – 1f(x)=3x4−5x2+7x−1 is f′′(x)=36×2−10f”(x) = 36x^2 – 10f′′(x)=36x2−10.

Third and Higher Derivatives:

Higher derivatives follow the same principle. For example, the third derivative f′′′(x)f”'(x)f′′′(x) is the derivative of the second derivative.

4. Applications of Differentiation

Differentiation is widely used in various fields to model rates of change. Here are some key applications:

a. Velocity and Acceleration

In physics, the velocity of an object is the derivative of its position function with respect to time, and acceleration is the derivative of velocity.

• If $s(t)$ is the position of an object at time $t$, then the velocity is v(t)=dsdtv(t) = \frac{ds}{dt}v(t)=dtds​.
• The acceleration is the rate of change of velocity, a(t)=dvdta(t) = \frac{dv}{dt}a(t)=dtdv​.

b. Optimization Problems

Differentiation is used to find maximum or minimum values of a function, which is essential in optimization problems (e.g., minimizing cost or maximizing profit). The first derivative tells us where the function’s slope is zero, which corresponds to possible maximum or minimum points.

Example 4:

Find the maximum or minimum of the function f(x)=−x2+4x+6f(x) = -x^2 + 4x + 6f(x)=−x2+4x+6.

Solution:

1. Find the first derivative:

f′(x)=−2x+4f'(x) = -2x + 4f′(x)=−2x+4

2. Set f′(x)=0f'(x) = 0f′(x)=0 to find the critical points:

$$-2x+4=0⟹x=2$$

3. To confirm whether it is a maximum or minimum, find the second derivative:

f′′(x)=−2f”(x) = -2f′′(x)=−2

Since f′′(x)=−2<0f”(x) = -2 < 0f′′(x)=−2<0, the function has a maximum at $x=2$.

Thus, the maximum value occurs at $x=2$.

Integration and Evaluation of Simple Algebraic Functions: A Comprehensive Guide

Integration is a fundamental concept in calculus that focuses on finding the accumulated total of a quantity. This process is especially important in solving problems related to areas under curves, volumes of solids, and rates of change. In this article, we will provide a detailed explanation of integration, its methods, and applications, specifically focusing on simple algebraic functions. By the end, you will understand how integration works and how to apply it to solve various mathematical problems.

What is Integration?

Integration is essentially the reverse of differentiation. While differentiation deals with finding the rate of change of a function, integration focuses on finding the accumulated or total change over a given interval. It helps calculate areas under curves, total distance traveled, and other quantities that accumulate over time.

Mathematically, the integral of a function represents the area under the graph of that function between two points on the x-axis. For example, finding the integral of the function f(x)=x2f(x) = x^2f(x)=x2 would give us the area under the curve from a starting point to an endpoint.

Basic Integration Formulas

Before diving into integration techniques, let’s review some basic integration rules for simple algebraic functions. These formulas are the building blocks of integration.

• Power Rule for Integration:
  If f(x)=xnf(x) = x^nf(x)=xn, where n≠−1n \neq -1n=−1, the integral is:∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1​+CExample:

  ∫x2dx=x33+C\int x^2 dx = \frac{x^3}{3} + C∫x2dx=3x3​+CHere, $C$ is the constant of integration.

• Integral of a Constant:
  If $f(x)=a$, where $a$ is a constant, the integral is:$$\int adx=ax+C$$Example:

  $$\int 5dx=5x+C$$

• Integral of Exponential Functions:
  If f(x)=exf(x) = e^xf(x)=ex, the integral is:∫exdx=ex+C\int e^x dx = e^x + C∫exdx=ex+C
• Integral of Trigonometric Functions:
  • $\int \sin (x)dx=-\cos (x)+C$
  • $\int \cos (x)dx=\sin (x)+C$

Methods of Integration

Now that we’ve covered the basic integration formulas, let’s dive into the different methods of integration used to evaluate more complex algebraic functions.

Substitution Method

The substitution method is used to simplify an integral by changing variables. This method is particularly useful when dealing with composite functions, where one function is nested inside another.

To use the substitution method, we follow these steps:

1. Identify a part of the integrand that can be substituted with a new variable.
2. Differentiate the substituted expression to find $du$.
3. Replace the original expression with the new variable and integrate.

Example:
Evaluate the integral ∫x⋅cos⁡(x2)dx\int x \cdot \cos(x^2) dx∫x⋅cos(x2)dx.

Let’s use substitution:

1. Let u=x2u = x^2u=x2. Therefore, $du=2xdx$, or du2=xdx\frac{du}{2} = x dx2du​=xdx.
2. Substituting into the integral, we get: ∫x⋅cos⁡(x2)dx=12∫cos⁡(u)du\int x \cdot \cos(x^2) dx = \frac{1}{2} \int \cos(u) du∫x⋅cos(x2)dx=21​∫cos(u)du
3. Now integrate $\cos (u)$: 12sin⁡(u)+C\frac{1}{2} \sin(u) + C21​sin(u)+C
4. Substitute back u=x2u = x^2u=x2: 12sin⁡(x2)+C\frac{1}{2} \sin(x^2) + C21​sin(x2)+C

Partial Fraction Decomposition Method

The partial fraction method is used when integrating rational functions (fractions where both the numerator and denominator are polynomials). The goal is to express the rational function as a sum of simpler fractions that are easier to integrate.

1. Factor the denominator into irreducible factors.
2. Express the rational function as a sum of fractions with unknown coefficients.
3. Solve for the coefficients and then integrate each term separately.

Example:
Evaluate the integral ∫1×2−1dx\int \frac{1}{x^2 – 1} dx∫x2−11​dx.

1. Factor the denominator: 1×2−1=1(x−1)(x+1)\frac{1}{x^2 – 1} = \frac{1}{(x – 1)(x + 1)}x2−11​=(x−1)(x+1)1​
2. Decompose the fraction: 1(x−1)(x+1)=Ax−1+Bx+1\frac{1}{(x – 1)(x + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1}(x−1)(x+1)1​=x−1A​+x+1B​
3. Solve for $A$ and $B$: $$1=A(x+1)+B(x-1)$$ Set $x=1$ and $x=-1$ to find A=12A = \frac{1}{2}A=21​ and B=−12B = \frac{-1}{2}B=2−1​.
4. Integrate each term: ∫1×2−1dx=12ln⁡∣x−1∣−12ln⁡∣x+1∣+C\int \frac{1}{x^2 – 1} dx = \frac{1}{2} \ln|x – 1| – \frac{1}{2} \ln|x + 1| + C∫x2−11​dx=21​ln∣x−1∣−21​ln∣x+1∣+C Simplifying: ∫1×2−1dx=12ln⁡∣x−1x+1∣+C\int \frac{1}{x^2 – 1} dx = \frac{1}{2} \ln \left| \frac{x – 1}{x + 1} \right| + C∫x2−11​dx=21​ln​x+1x−1​​+C

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. It’s typically used when the integrand is the product of two functions.

The formula for integration by parts is:

$$\int udv=uv-\int vdu$$

Where $u$ and $v$ are differentiable functions. The goal is to choose $u$ and $dv$ such that the resulting integral is easier to solve.

Example:
Evaluate the integral ∫xexdx\int x e^x dx∫xexdx.

1. Let $u=x$ and dv=exdxdv = e^x dxdv=exdx.
2. Then, $du=dx$ and v=exv = e^xv=ex.
3. Apply the integration by parts formula: ∫xexdx=xex−∫exdx\int x e^x dx = x e^x – \int e^x dx∫xexdx=xex−∫exdx
4. Integrate: xex−ex+Cx e^x – e^x + Cxex−ex+C

Applications of Integration

Integration has several practical applications, including:

• Finding the area under a curve: By integrating a function, you can calculate the total area under the curve between two points.
• Physics: Integration is used to calculate quantities like distance traveled, velocity, and acceleration.
  • Finding Areas: The most common application of integration is finding the area under curves. For instance, in physics, the area under a velocity-time graph gives the total distance traveled.
  • Calculating Volume: By rotating a function around an axis, integration can be used to find the volume of a solid of revolution.

Accumulation Problems: Integration can model situations where a quantity is continuously accumulated over time, such as the growth of a population or the accumulation of interest in finance.Economics: It helps in finding the total cost, revenue, or profit over a given time period.

Integration and Evaluation of Simple Algebraic Functions

Integration is a fundamental concept in calculus, representing the reverse of differentiation. It is used to find areas under curves, volumes of solids, and to solve a variety of problems involving accumulation and growth. In this lesson, we will focus on integrating simple algebraic functions, which are typically polynomials or rational functions. We will also explore how to evaluate definite and indefinite integrals, and apply them to real-world problems.

1. Understanding Integration

Integration can be viewed as the process of finding the antiderivative of a function. The antiderivative of a function is another function whose derivative is the original function.

Two Types of Integrals:

1. Indefinite Integral: The indefinite integral represents the family of antiderivatives of a function. It does not have specific limits, and the result always includes a constant of integration, denoted by $C$.The general form is:

   $$\int f(x)dx=F(x)+C$$where $F(x)$ is the antiderivative of $f(x)$.

2. Definite Integral: The definite integral calculates the total accumulation (area under the curve) of the function between two specific limits, $a$ and $b$.The general form is:

   ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) – F(a)∫ab​f(x)dx=F(b)−F(a)where $F(x)$ is the antiderivative of $f(x)$, and $a$ and $b$ are the limits of integration.

2. Basic Integration Rules

Here are the basic integration rules that will help you solve simple algebraic functions:

• Power Rule: If f(x)=xnf(x) = x^nf(x)=xn, where n≠−1n \neq -1n=−1, then:∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1​+CExample:

  ∫x3 dx=x44+C\int x^3 \, dx = \frac{x^4}{4} + C∫x3dx=4x4​+C

• Constant Rule: If $f(x)=c$, where $c$ is a constant, then:$$\int cdx=cx+C$$Example:

  $$\int 5dx=5x+C$$

• Sum Rule: The integral of a sum of functions is the sum of the integrals:$$\int (f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$$
• Constant Multiple Rule: If a constant $c$ is multiplied by the function, you can factor it out:$$\int c\cdot f(x)dx=c\cdot \int f(x)dx$$

3. Examples of Simple Algebraic Integration

Let’s go through some examples to better understand integration.

Example 1: Integrating a Polynomial Function

Integrate f(x)=3×2+5x+2f(x) = 3x^2 + 5x + 2f(x)=3x2+5x+2.

Solution:
Apply the sum and power rules for each term:

∫(3×2+5x+2) dx=∫3×2 dx+∫5x dx+∫2 dx\int (3x^2 + 5x + 2) \, dx = \int 3x^2 \, dx + \int 5x \, dx + \int 2 \, dx∫(3x2+5x+2)dx=∫3x2dx+∫5xdx+∫2dx

Now, use the power rule for each term:

∫3×2 dx=3×33=x3\int 3x^2 \, dx = \frac{3x^3}{3} = x^3∫3x2dx=33x3​=x3 ∫5x dx=5×22\int 5x \, dx = \frac{5x^2}{2}∫5xdx=25x2​ $$\int 2dx=2x$$

Thus, the integral is:

∫(3×2+5x+2) dx=x3+5×22+2x+C\int (3x^2 + 5x + 2) \, dx = x^3 + \frac{5x^2}{2} + 2x + C∫(3x2+5x+2)dx=x3+25x2​+2x+CExample 2: Integrating a Rational Function

Integrate f(x)=1xf(x) = \frac{1}{x}f(x)=x1​.

Solution:
This is a special case where the integral is known:

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln|x| + C∫x1​dx=ln∣x∣+C4.

Definite Integration

Definite integration allows us to compute the total accumulated value of a function over a given interval, represented by limits $a$ and $b$. It can be used to calculate areas under curves.

Example 3:

Find the area under the curve $f(x)=2x+1$ from $x=0$ to $x=3$.

Solution:

1. First, find the indefinite integral: ∫(2x+1) dx=x2+x+C\int (2x + 1) \, dx = x^2 + x + C∫(2x+1)dx=x2+x+C
2. Now, apply the limits of integration $a=0$ and $b=3$: ∫03(2x+1) dx=[x2+x]03=(32+3)−(02+0)\int_0^3 (2x + 1) \, dx = \left[ x^2 + x \right]_0^3 = (3^2 + 3) – (0^2 + 0)∫03​(2x+1)dx=[x2+x]03​=(32+3)−(02+0) $$=(9+3)-(0)=12$$

Thus, the area under the curve from $x=0$ to $x=3$ is 12 square units.

5. Applications of Integration

Integration has wide-ranging applications in various fields, especially in physics, engineering, economics, and biology. Some practical uses include:

Finding Areas: The most common application of integration is finding the area under curves. For instance, in physics, the area under a velocity-time graph gives the total distance traveled.

Calculating Volume: By rotating a function around an axis, integration can be used to find the volume of a solid of revolution.

Accumulation Problems: Integration can model situations where a quantity is continuously accumulated over time, such as the growth of a population or the accumulation of interest in finance.