Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Integration of simple Algebraic functions
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Integration of simple Algebraic functions
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Subject: General Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 6
Theme: Introductory Calculus
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Recognize in tegrationas the reverse of differentiation Recognize somestandard in tegrals of polymids and algebraictractions Apply sometechniques of integration such asa) in tegration by substitutionb) in tegration by partsc) in tegration by partial fractions Apply in tegration to real life situation.
This section outlines the core principles and methods for integrating simple algebraic functions. 2.
1. Introduction to Integration: The Reverse of Differentiation Integration, also known as antidifferentiation, is the process of finding a function whose derivative is a given function. If the derivative of $F(x)$ is $f(x)$, then $F(x)$ is an antiderivative of $f(x)$.
Notation: The integral of a function $f(x)$ with respect to $x$ is denoted by $\int f(x) dx$.
Constant of Integration: If $F(x)$ is an antiderivative of $f(x)$, then $F(x) + C$ is also an antiderivative, where $C$ is an arbitrary constant called the constant of integration. This is because the derivative of any constant is zero. Hence, $\frac{d}{dx}(F(x) + C) = F'(x) + 0 = f(x)$. 2.
2. Standard Integrals of Polynomials and Algebraic Fractions Power Rule for Integration: For any real number $n \neq -1$: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ Example 1: Integrate $x^3$. $\int x^3 dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$ Integral of a Constant: $\int k dx = kx + C$, where $k$ is a constant.
Example 2: Integrate $5$. $\int 5 dx = 5x + C$ Integral of $1/x$: For $n=-1$, the power rule does not apply. $\int \frac{1}{x} dx = \ln|x| + C$ This is because $\frac{d}{dx}(\ln|x|) = \frac{1}{x}$. Rules for Sums, Differences, and Scalar Multiples: $\int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx$ $\int k f(x) dx = k \int f(x) dx$ Example 3: Integrate $3x^2 - \frac{2}{x} + 7$. $\int (3x^2 - \frac{2}{x} + 7) dx = 3 \int x^2 dx - 2 \int \frac{1}{x} dx + \int 7 dx$ $= 3\left(\frac{x^{2+1}}{2+1}\right) - 2(\ln|x|) + 7x + C$ $= 3\left(\frac{x^3}{3}\right) - 2\ln|x| + 7x + C$ $= x^3 - 2\ln|x| + 7x + C$ 2.
3. Techniques of Integration a) Integration by Substitution (U-Substitution) This technique simplifies an integral by transforming it into a simpler form using a new variable. It is effective when the integrand contains a function and its derivative.
Steps:
1. Choose a substitution $u = g(x)$, typically the inner function or a part that simplifies the integral.
2. Calculate $\frac{du}{dx}$ and express $dx$ in terms of $du$ (i.e., $dx = \frac{du}{g'(x)}$ or $dx = \frac{1}{g'(x)} du$).
3. Substitute $u$ and $dx$ into the integral to express it entirely in terms of $u$.
4. Integrate the new expression with respect to $u$.
5. Substitute back $g(x)$ for $u$ to express the result in terms of the original variable $x$.
Example 4: Integrate $\int (2x+3)^4 dx$.
1. Let $u = 2x+3$. 2. $\frac{du}{dx} = 2 \implies dx = \frac{1}{2} du$.
3. Substitute: $\int u^4 \left(\frac{1}{2} du\right) = \frac{1}{2} \int u^4 du$.
4. Integrate: $\frac{1}{2} \left(\frac{u^5}{5}\right) + C = \frac{u^5}{10} + C$.
5. Substitute back: $\frac{(2x+3)^5}{10} + C$. Example 5 (Nigerian Context - Revenue Rate): A company's marginal revenue (rate of change of revenue) for selling $x$ units of a product (e.g., bags of cement) is given by $MR = 500(2x+1)^2$ Naira per unit. Find the total revenue function. Total Revenue $R(x) = \int MR \, dx = \int 500(2x+1)^2 dx$.
1. Let $u = 2x+1$. 2. $\frac{du}{dx} = 2 \implies dx = \frac{1}{2} du$.
3. Substitute: $\int 500 u^2 \left(\frac{1}{2} du\right) = 250 \int u^2 du$.
4. Integrate: $250 \left(\frac{u^3}{3}\right) + C = \frac{250u^3}{3} + C$.
5. Substitute back: $R(x) = \frac{250(2x+1)^3}{3} + C$. (The constant $C$ would typically be zero if no revenue is generated when zero units are sold). b) Integration by Parts This technique is used to integrate products of functions and is derived from the product rule for differentiation.
The formula is: $\int u \, dv = uv - \int v \, du$ Steps:
1. Carefully choose $u$ and $dv$ from the integrand. A common mnemonic for choosing $u$ is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) – choose the function that comes first in this order. The remaining part is $dv$.
2. Differentiate $u$ to find $du$.
3. Integrate $dv$ to find $v$.
4. Substitute $u, v, du, dv$ into the integration by parts formula.
5. Evaluate the new integral $\int v \, du$. * product rule for differentiation.
The formula is: $\int u \, dv = uv - \int v \, du$ Steps:
1. Carefully choose $u$ and $dv$ from the integrand. A common mnemonic for choosing $u$ is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) – choose the function that comes first in this order. The remaining part is $dv$.
2. Differentiate $u$ to find $du$.
3. Integrate $dv$ to find $v$.
4. Substitute $u, v, du, dv$ into the integration by parts formula.
5. Evaluate the new integral $\int v \, du$.
Example 6 (Algebraic functions): Integrate $\int x(x+1)^3 dx$. (This can also be done by substitution, but demonstrates parts)
1. Let $u = x$ (Algebraic) and $dv = (x+1)^3 dx$. 2. $du = 1 \, dx$. 3. $v = \int (x+1)^3 dx = \frac{(x+1)^4}{4}$. (Using substitution $w=x+1$).
4. Apply formula: $\int x(x+1)^3 dx = x \cdot \frac{(x+1)^4}{4} - \int \frac{(x+1)^4}{4} \cdot 1 \, dx$.
5. Evaluate $\int v \, du$: $\frac{x(x+1)^4}{4} - \frac{1}{4} \int (x+1)^4 dx$ $= \frac{x(x+1)^4}{4} - \frac{1}{4} \left(\frac{(x+1)^5}{5}\right) + C$ $= \frac{x(x+1)^4}{4} - \frac{(x+1)^5}{20} + C$.
Example 7 (Algebraic and Logarithmic): Integrate $\int x \ln x \, dx$.
1. According to LIATE, $u = \ln x$ (Logarithmic) and $dv = x \, dx$ (Algebraic). 2. $du = \frac{1}{x} \, dx$. 3. $v = \int x \, dx = \frac{x^2}{2}$.
4. Apply formula: $\int x \ln x \, dx = \ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$.
5. Evaluate $\int v \, du$: $\frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx$ $= \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx$ $= \frac{x^2}{2} \ln x - \frac{1}{2} \left(\frac{x^2}{2}\right) + C$ $= \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$. c) Integration by Partial Fractions This technique is used to integrate rational functions (fractions where the numerator and denominator are polynomials) by decomposing them into a sum of simpler fractions that are easier to integrate. This is typically applied when the degree of the numerator is less than the degree of the denominator.
Steps:
1. Factorize the denominator of the rational function.
2. Decompose the rational function into partial fractions based on the factors of the denominator. For a distinct linear factor $(ax+b)$, the partial fraction is $\frac{A}{ax+b}$. For a repeated linear factor $(ax+b)^n$, the partial fractions are $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$. (For SS3, focus mainly on distinct linear factors).
3. Solve for the unknown constants ($A, B, C, \dots$) by equating coefficients or by substituting specific values of $x$.
4. Integrate each partial fraction using standard integration rules (mostly $\int \frac{1}{x} dx = \ln|x| + C$).
Example 8: Integrate $\int \frac{1}{x^2-1} dx$.
1. Factorize denominator: $x^2-1 = (x-1)(x+1)$.
2. Decompose: $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$.
3. Clear denominators: $1 = A(x+1) + B(x-1)$. To find $A$, set $x=1$: $1 = A(1+1) + B(1-1) \implies 1 = 2A \implies A = \frac{1}{2}$. To find $B$, set $x=-1$: $1 = A(-1+1) + B(-1-1) \implies 1 = -2B \implies B = -\frac{1}{2}$.
4. Substitute constants and integrate: $\int \left(\frac{1/2}{x-1} - \frac{1/2}{x+1}\right) dx = \frac{1}{2} \int \frac{1}{x-1} dx - \frac{1}{2} \int \frac{1}{x+1} dx$ $= \frac{1}{2} \ln|x-1| - \frac{1}{2} \ln|x+1| + C$ $= \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right| + C$. Example 9 (Nigerian Context - Production Rate): The rate of production of a commodity (e.g., textiles) at a factory in Aba, Nigeria, can be modeled by $\frac{2}{x^2+2x} $ (in thousands of units per month), where $x$ is the number of months since production started. Find the total cumulative production after $x$ months. Total production $P(x) = \int \frac{2}{x^2+2x} dx$.
1. Factorize denominator: $x^2+2x = x(x+2)$.
2. Decompose: $\frac{2}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2}$.
3. Clear denominators: $2 = A(x+2) + Bx$. To find $A$, set $x=0$: $2 = A(0+2) + B(0) \implies 2 = 2A \implies A = 1$. * To find $B$, set $x=-2$: $2 = A(-2+2) + B(-2) \implies 2 = -2B \implies B = -1$.
4. Substitute constants and integrate: $\int \left(\frac{1}{x} - \frac{1}{x+2}\right) dx = \int \frac{1}{x} dx - \int \frac{1}{x+2} dx$ $= \ln|x| - \ln|x+2| + C$ $= \ln\left|\frac{x}{x+2}\right| + C$. (Assuming production starts from zero, $C$ Factorize denominator: $x^2+2x = x(x+2)$.
2. Decompose: $\frac{2}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2}$.
3. Clear denominators: $2 = A(x+2) + Bx$. To find $A$, set $x=0$: $2 = A(0+2) + B(0) \implies 2 = 2A \implies A = 1$. To find $B$, set $x=-2$: $2 = A(-2+2) + B(-2) \implies 2 = -2B \implies B = -1$.
4. Substitute constants and integrate: $\int \left(\frac{1}{x} - \frac{1}{x+2}\right) dx = \int \frac{1}{x} dx - \int \frac{1}{x+2} dx$ $= \ln|x| - \ln|x+2| + C$ $= \ln\left|\frac{x}{x+2}\right| + C$. (Assuming production starts from zero, $C$ would be determined based on initial conditions). d) Application of Integration to Real-life Situations Integration is applied to find: Area under a curve: If $f(x)$ represents a rate (e.g., velocity, marginal cost), $\int f(x) dx$ represents the total accumulation (e.g., total distance, total cost).
Total Change/Accumulation: If $f(t)$ is a rate of change, $\int_a^b f(t) dt$ gives the net change in the quantity from $t=a$ to $t=b$.
Economics: Total cost from marginal cost, total revenue from marginal revenue, consumer and producer surplus.
Physics: Displacement from velocity, velocity from acceleration.
Biology: Population growth, spread of diseases.
Engineering: Volume of material, work done by a force. * Capital Market Issues: Calculating the total return on an investment with continuously varying interest rates, or continuous compounding. For example, if an investment grows at a rate $r(t)$, the total accumulated value over a period can be found by integrating. --- Phase 1: Introduction (15 minutes)
Teacher Activity: Begin by reviewing differentiation: Ask students to differentiate simple polynomial functions (e.g., $x^3$, $4x^2-5x$). Introduce the concept of working backwards: "If we know the derivative, can we find the original function?" Define integration as the reverse process of differentiation (antidifferentiation). Explain the integral symbol ($\int$), integrand, and the importance of the constant of integration ($C$). Illustrate with simple examples like $\frac{d}{dx}(x^2) = 2x$, so $\int 2x dx = x^2 + C$.
Briefly state the overarching goal: to apply integration to solve practical problems.
Student Activity: Participate in the review of differentiation, solving problems on the board or in their notebooks. Listen attentively to the introduction of integration, taking notes. Ask clarifying questions about the concept of antiderivative and the constant of integration.
Phase 2: Development of Concepts and Techniques (60 minutes)
Teacher Activity: 2.
1. Standard Integrals (20 minutes): Introduce the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1} + C$) with clear steps and examples. Emphasize $n \neq -1$. Demonstrate integration of constants ($\int k dx = kx + C$). Explain the special case for $\int \frac{1}{x} dx = \ln|x| + C$. Work through examples combining these rules (sums, differences, scalar multiples). Provide practice problems for students to solve individually or in pairs. 2.
2. Integration by Substitution (15 minutes): Explain the rationale for substitution: simplifying complex integrands. Walk through the steps of U-substitution clearly, using Example 4 and 5 (Nigerian context) as demonstration. Emphasize choosing the correct $u$ and expressing $dx$ in terms of $du$. Supervise students as they try a similar problem. 2.
3. Integration by Parts (15 minutes): Introduce the formula $\int u \, dv = uv - \int v \, du$. Explain the LIATE rule (or similar heuristic) for choosing $u$ and $dv$. Work through Example 6 and 7, emphasizing the step-by-step process of finding $u, dv, du, v$. Provide a short practice problem. 2.
4. Integration by Partial Fractions (10 minutes): Explain when to use partial fractions (rational functions). Focus on decomposing functions with distinct linear factors in the denominator. Demonstrate with Example 8 and 9 (Nigerian context), showing how to find constants $A, B$. Guide students through the integration of the resulting simple fractions.
Student Activity: Actively participate in solving standard integral problems, practicing the power rule, constant rule, and $1/x$ rule. Work through assigned problems for standard integrals. Follow the teacher's explanation of integration by substitution, attempting guided practice problems. Understand the LIATE rule for integration by parts and practice identifying $u$ and $dv$. Learn the steps for decomposing rational functions into partial fractions and integrating them. Ask questions for clarification on any challenging step or concept.
Phase 3: Real-life Applications and Conclusion (15 minutes)
Teacher Activity: Discuss various real-life applications of integration, linking back to the Nigerian context. Elaborate on how integration helps determine total revenue from marginal revenue, total production from production rate (as in Example 5 and 9). Discuss its relevance in the capital market for understanding accumulated wealth or total investment returns over time. Summarize the key concepts and techniques covered: standard integrals, substitution, parts, partial fractions. Address any remaining questions.
Student Activity: Engage in discussions about the real-life applications of integration, providing examples from their environment or current events. Consolidate their understanding of the various integration techniques. Participate in a final Q&A session.
Teaching Aids: Whiteboard/Blackboard, markers/chalk, prepared examples and exercises, charts illustrating integral formulas.
Classroom Management: Group work for practice, peer tutoring, individual problem-solving, monitoring student progress and offering individual support. ---
Estimating Resource Depletion (e.g., Crude Oil Reserves): In the Niger Delta, if the rate of crude oil extraction from a particular oil field is known as a function of time, integration can be used to calculate the total amount of oil extracted over a certain period. This helps policymakers and oil companies in Nigeria manage resources and forecast depletion. For instance, if the extraction rate is $R(t)$ barrels/day, $\int_{t_1}^{t_2} R(t) dt$ gives the total barrels extracted between $t_1$ and $t_2$. Infrastructure Development (e.g., Road Construction): When constructing a road in Nigeria, engineers can use integration to calculate the total amount of materials (e.g., asphalt, concrete) needed for a section of varying width or thickness. If the cross-sectional area of the road varies along its length, integrating the area function with respect to length gives the total volume of material required, thus aiding in budgeting and logistics for projects like the Lagos-Calabar Coastal Highway. Financial Planning in the Nigerian Capital Market: An investor contributing to a mutual fund or fixed deposit account with continuous compounding or varying interest rates over time can use integration to determine the total accumulated value of their investment. If the instantaneous rate of change of investment value is $V'(t)$, then the total value after $T$ years is $\int_0^T V'(t) dt$. This is particularly useful for understanding the long-term growth of pensions or savings in the Nigerian financial system. For example, if the rate of growth of an investment in a government bond is $k(t)P(t)$ where $P(t)$ is the principal at time $t$, integration helps calculate the total return. ---