Integration Pt. 2 - Fundamental Theorems

## Area Under a Curve & Riemann Sums ### Approximation & Limit Definition **Riemann Sums** serve as the foundational method for approximating the area bounded by a function curve and the x-axis. By partitioning an interval into sub-intervals and constructing geometric shapes (rectangles or trapezoids), we approximate the total area. This concept bridges pre-calculus geometry and calculus. The **Definite Integral** is rigorously defined as the limit of a Riemann Sum as the number of sub-intervals ($n$) approaches infinity (or as the width of the sub-intervals, $\Delta x$, approaches zero). $$ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x $$ ### Summation Methods, Error Analysis, & Nuances The approximation method depends on where the height of the rectangle is sampled within the sub-interval. * **LRAM (Left Riemann Sum):** Height is $f(x)$ at the left endpoint. * **Nuance:** Underestimates if $f(x)$ is **increasing**. Overestimates if $f(x)$ is **decreasing**. * **RRAM (Right Riemann Sum):** Height is $f(x)$ at the right endpoint. * **Nuance:** Overestimates if $f(x)$ is **increasing**. Underestimates if $f(x)$ is **decreasing**. * **MRAM (Midpoint Riemann Sum):** Height is $f(x)$ at the midpoint. * **Accuracy:** Generally more accurate than LRAM or RRAM but requires calculating midpoint coordinates. * **Trapezoidal Rule:** Uses trapezoids connecting points $(x_i, f(x_i))$ and $(x_{i+1}, f(x_{i+1}))$. * **Formula:** $Area \approx \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + ... + f(x_n)]$. * **Error Analysis (Concavity):** The error depends on **concavity**, not slope. * Concave **Up**: Overestimates (secant lines lie above curve). * Concave **Down**: Underestimates (secant lines lie below curve). **Common Pitfalls:** * Assuming RRAM always overestimates (it depends on slope, not concavity). * Confusing "Total Area" (geometric area) with "Net Area" (integral value). Riemann sums calculate net signed area. * Unequal sub-intervals: Standard formulas assume equal $\Delta x$, but AP table problems often use unequal widths. You must calculate the area of each shape individually. ### Worked Examples: LRAM and Trapezoidal Rule **Scenario:** Estimate the area under $f(x) = x^2 + 2$ on $[0, 6]$ using 3 equal sub-intervals. graph[x^2 + 2][-1][7] **1. Setup Interval Width ($\Delta x$)** $$ \Delta x = \frac{b-a}{n} = \frac{6-0}{3} = 2 $$ Intervals: $[0, 2], [2, 4], [4, 6]$ **2. LRAM Calculation (Left Endpoints: 0, 2, 4)** * Calculate heights: $$f(0) = 0^2 + 2 = 2$$ $$f(2) = 2^2 + 2 = 6$$ $$f(4) = 4^2 + 2 = 18$$ * Sum areas (Width $\times$ Height): $$Area \approx 2(2) + 2(6) + 2(18)$$ $$Area \approx 4 + 12 + 36 = 52$$ * *Analysis:* Since $x^2+2$ is increasing, LRAM ($52$) is an **underestimate**. **3. Trapezoidal Approximation** * Use endpoints: $0, 2, 4, 6$. $$f(0)=2, f(2)=6, f(4)=18, f(6)=38$$ * Apply formula $\frac{\Delta x}{2}(y_0 + 2y_1 + ... + y_n)$: $$Area \approx \frac{2}{2} [2 + 2(6) + 2(18) + 38]$$ $$Area \approx 1 [2 + 12 + 36 + 38]$$ $$Area \approx 88$$ * *Analysis:* Since $x^2+2$ is concave up ($f''(x)=2 > 0$), Trapezoidal ($88$) is an **overestimate**. --- ## Definite Integrals & Properties ### Signed Area & Formal Definition A **Definite Integral**, denoted $\int_a^b f(x) \, dx$, represents the **net signed area** bounded by the curve and the x-axis from $x=a$ to $x=b$. * Area **above** the x-axis is positive. * Area **below** the x-axis is negative. This distinction is vital for interpreting physical contexts (e.g., displacement vs. distance). ### Essential Properties & Algebraic Rules Mastery of these manipulations is required for "abstract function" problems where no explicit equation is given. * **Linearity:** $\int_a^b [kf(x) \pm g(x)] \, dx = k\int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx$ * **Additivity (Interval Addition):** $\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx$ * **Reversing Limits:** $\int_b^a f(x) \, dx = - \int_a^b f(x) \, dx$ * **Zero Width:** $\int_a^a f(x) \, dx = 0$ * **Dominance:** If $f(x) \ge g(x)$ on $[a, b]$, then $\int_a^b f(x) \, dx \ge \int_a^b g(x) \, dx$. **Niche Application (Geometric Evaluation):** AP Exams often require evaluating integrals using geometry rather than antiderivatives. * $\sqrt{r^2 - x^2}$ represents a semi-circle. * $|x-c|$ represents a 'V' shape (triangles). ### Worked Example: Properties & Geometry **Given:** $\int_0^5 f(x) dx = 10$ and $\int_0^7 f(x) dx = 13$. **Find:** $\int_5^7 [2f(x) + 3] dx$. **1. Break down the target integral using Linearity** $$ 2\int_5^7 f(x) dx + \int_5^7 3 dx $$ **2. Solve for $\int_5^7 f(x) dx$ using Additivity** $$ \int_0^7 f(x) dx = \int_0^5 f(x) dx + \int_5^7 f(x) dx $$ $$ 13 = 10 + \int_5^7 f(x) dx $$ $$ \int_5^7 f(x) dx = 3 $$ **3. Evaluate the constant integral $\int_5^7 3 dx$** * *Method:* Geometry (Rectangle with height 3, width $7-5=2$). $$ 3 \times (7-5) = 6 $$ **4. Combine Results** $$ 2(3) + 6 $$ $$ 6 + 6 = 12 $$ --- ## The Fundamental Theorem of Calculus (Evaluation) ### Connection Between Derivatives & Integrals Often called the **Evaluation Theorem** or FTC Part 1 in some texts, this theorem provides the practical method for evaluating definite integrals without Riemann sums. It connects differential and integral calculus: differentiation and integration are inverse operations. $$ \text{If } F'(x) = f(x), \text{ then } \int_a^b f(x) \, dx = F(b) - F(a) $$ ### Requirements & Notation * **Continuity:** $f(x)$ must be continuous on $[a, b]$. * **Antiderivative ($F(x)$):** You must be able to find a function $F$ such that $F' = f$. * **Notation:** The bracket notation is standard: $[F(x)]_a^b = F(b) - F(a)$. **Common Pitfalls:** * **Forgetting the Chain Rule (Reverse):** When integrating $e^{3x}$, students often write $e^{3x}$ instead of $\frac{1}{3}e^{3x}$. * **Discontinuities:** You cannot apply FTC directly across a vertical asymptote (e.g., $\int_{-1}^1 \frac{1}{x^2} dx$ diverges; blindly applying power rule yields a negative result, which is impossible for a positive function). ### Worked Example: Definite Evaluation **Evaluate:** $\int_1^2 (3x^2 - \frac{4}{x}) \, dx$ graph[3x^2 - 4/x][0.5][3] **1. Find the Antiderivative** * Recall Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1}$ * Recall Log Rule: $\int \frac{1}{x} dx = \ln|x|$ $$ F(x) = 3\left(\frac{x^3}{3}\right) - 4\ln|x| = x^3 - 4\ln|x| $$ **2. Apply FTC ($F(b) - F(a)$)** $$ [x^3 - 4\ln|x|]_1^2 $$ $$ = (2^3 - 4\ln(2)) - (1^3 - 4\ln(1)) $$ **3. Simplify** * Note: $\ln(1) = 0$ $$ = (8 - 4\ln 2) - (1 - 0) $$ $$ = 7 - 4\ln 2 $$ --- ## The Second Fundamental Theorem (Accumulation) ### Derivative of an Accumulation Function This theorem handles functions defined *as* integrals. It states that if you integrate a function and then differentiate it, you return to the original function (with variable adjustments). This models rate-of-change problems where the quantity is defined by accumulated area. $$ \frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x) $$ ### Chain Rule Extension (Crucial) If the upper limit is a function $g(x)$ rather than just $x$, the Chain Rule applies. This is the most common version seen in FRQs. $$ \frac{d}{dx} \left[ \int_a^{g(x)} f(t) \, dt \right] = f(g(x)) \cdot g'(x) $$ **Key Components:** * **Bottom Limit Constant:** The lower limit $a$ must be a constant. If it is a variable, use properties to split the integral. * **Variable Swap:** The variable $t$ inside the integral is a "dummy variable." It gets replaced by the expression in the upper limit. ### Worked Example: FTC with Chain Rule **Problem:** Let $H(x) = \int_{\pi}^{x^3} \cos(t) \, dt$. Find $H'(x)$. **1. Identify Structure** * Upper limit is a function $g(x) = x^3$. * Integrand is $f(t) = \cos(t)$. * Formula: $f(g(x)) \cdot g'(x)$. **2. Differentiate Upper Limit** $$ g(x) = x^3 \implies g'(x) = 3x^2 $$ **3. Substitute and Multiply** * Replace $t$ with $x^3$: $\cos(x^3)$ * Multiply by derivative: $\cos(x^3) \cdot 3x^2$ **Final Answer:** $$ H'(x) = 3x^2 \cos(x^3) $$ --- ## Average Value & MVT for Integrals ### Concept: "Leveling Out" the Area * **Average Value:** If you melted the area under the curve into a perfect rectangle with width $(b-a)$, what would the height be? That height is the average value. * **Mean Value Theorem (Integrals):** Guarantees that a continuous function actually *attains* its average value at least once in the interval. ### Formulas & Geometry **Average Value Formula:** $$ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx $$ **MVT for Integrals Statement:** If $f$ is continuous on $[a, b]$, there exists at least one number $c$ in $(a, b)$ such that: $$ f(c) = f_{avg} $$ Or equivalently: $\int_a^b f(x) \, dx = f(c)(b-a)$. **Nuance:** Distinguish between **Average Rate of Change** (Slope of secant: $\frac{f(b)-f(a)}{b-a}$) and **Average Value of a Function** (Integral height: $\frac{1}{b-a}\int f$). ### Worked Example: Finding 'c' **Problem:** Find the value $c$ guaranteed by the MVT for Integrals for $f(x) = 3x^2$ on $[0, 2]$. **1. Calculate the Integral** $$ \int_0^2 3x^2 \, dx = [x^3]_0^2 = 2^3 - 0^3 = 8 $$ **2. Calculate Average Value** $$ f_{avg} = \frac{1}{2-0} (8) = 4 $$ **3. Set $f(c) = f_{avg}$ and Solve** $$ 3c^2 = 4 $$ $$ c^2 = \frac{4}{3} $$ $$ c = \pm \frac{2}{\sqrt{3}} $$ **4. Check Interval $[0, 2]$** * Discard $-\frac{2}{\sqrt{3}}$ (outside interval). * Keep $c = \frac{2}{\sqrt{3}}$. --- ## Rectilinear Motion (Integration Applications) ### From Acceleration to Position Integration allows us to move "up" the ladder of motion: $$ a(t) \xrightarrow{\int} v(t) \xrightarrow{\int} s(t) $$ where $s$ is position, $v$ is velocity, and $a$ is acceleration. ### Displacement vs. Total Distance (High Priority) This is the single most common trap in particle motion problems. * **Displacement (Net Change):** How far you are from where you started. $$ \text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt = s(t_2) - s(t_1) $$ * **Total Distance Traveled:** The sum of all ground covered (forward and backward). Requires absolute value. $$ \text{Total Distance} = \int_{t_1}^{t_2} |v(t)| \, dt $$ **Current Position Formula:** To find the actual position $s(t)$ given an initial position $s(0)$: $$ s(t) = s(0) + \int_0^t v(x) \, dx $$ ### Worked Example: Motion Analysis **Given:** A particle moves with velocity $v(t) = t^2 - 4$ m/s. Initial position $s(0) = 5$. **Find:** 1. Displacement on $[0, 3]$. 2. Total Distance Traveled on $[0, 3]$. 3. Position at $t=3$. graph[x^2-4][-1][4] **1. Displacement** $$ \int_0^3 (t^2 - 4) \, dt = [\frac{t^3}{3} - 4t]_0^3 $$ $$ = (\frac{27}{3} - 12) - 0 $$ $$ = 9 - 12 = -3 \text{ meters} $$ **2. Total Distance (Requires splitting at zeros)** * Set $v(t) = 0$: $t^2 - 4 = 0 \implies t = 2$. * Function is negative on $[0, 2]$ and positive on $[2, 3]$. * Split integral and apply absolute value to negative regions: $$ \left| \int_0^2 (t^2 - 4) dt \right| + \int_2^3 (t^2 - 4) dt $$ * Evaluate Part 1 ($0$ to $2$): $$ [\frac{t^3}{3} - 4t]_0^2 = (\frac{8}{3} - 8) = \frac{8}{3} - \frac{24}{3} = -\frac{16}{3} $$ $$ \text{Absolute value} \to \frac{16}{3} $$ * Evaluate Part 2 ($2$ to $3$): $$ \text{Displacement on } [0,3] \text{ was } -3 \text{ (or } -\frac{9}{3}\text{)} $$ $$ \text{Since } \int_0^3 = \int_0^2 + \int_2^3 \implies -3 = -\frac{16}{3} + \int_2^3 $$ $$ \int_2^3 = -3 + \frac{16}{3} = \frac{7}{3} $$ * Total Distance: $\frac{16}{3} + \frac{7}{3} = \frac{23}{3} \text{ meters}$. **3. Position at $t=3$** $$ s(3) = s(0) + \text{Displacement} $$ $$ s(3) = 5 + (-3) = 2 $$

Area Under a Curve & Riemann SumsApproximation & Limit Definition Riemann Sums serve as the foundational method for approximating the area bounded by a function curve and the x-axis. By partitioning an interval into sub-intervals and constructing geometric shapes (rectangles or trapezoids), we approximate the total area.This concept bridges pre-calculus geometry and calculus. The Definite Integral is rigorously defined as the limit of a Riemann Sum as the number of sub-intervals () approaches infinity (or as the width of the sub-intervals, , approaches zero).Summation Methods, Error Analysis, & NuancesThe approximation method depends on where the height of the rectangle is sampled within the sub-interval.LRAM (Left Riemann Sum): Height is at the left endpoint.Nuance: Underestimates if is increasing. Overestimates if is decreasing.RRAM (Right Riemann Sum): Height is at the right endpoint.Nuance: Overestimates if is increasing. Underestimates if is decreasing.MRAM (Midpoint Riemann Sum): Height is at the midpoint.Accuracy: Generally more accurate than LRAM or RRAM but requires calculating midpoint coordinates.Trapezoidal Rule: Uses trapezoids connecting points and .Formula: .Error Analysis (Concavity): The error depends on concavity, not slope.Concave Up: Overestimates (secant lines lie above curve).Concave Down: Underestimates (secant lines lie below curve).Common Pitfalls:Assuming RRAM always overestimates (it depends on slope, not concavity).Confusing "Total Area" (geometric area) with "Net Area" (integral value). Riemann sums calculate net signed area.Unequal sub-intervals: Standard formulas assume equal , but AP table problems often use unequal widths. You must calculate the area of each shape individually.Worked Examples: LRAM and Trapezoidal RuleScenario: Estimate the area under on using 3 equal sub-intervals.1. Setup Interval Width ()Intervals: 2. LRAM Calculation (Left Endpoints: 0, 2, 4)Calculate heights:Sum areas (Width Height):Analysis: Since is increasing, LRAM () is an underestimate.3. Trapezoidal ApproximationUse endpoints: .Apply formula :Analysis: Since is concave up (), Trapezoidal () is an overestimate.Definite Integrals & PropertiesSigned Area & Formal DefinitionA Definite Integral, denoted , represents the net signed area bounded by the curve and the x-axis from to .Area above the x-axis is positive.Area below the x-axis is negative.This distinction is vital for interpreting physical contexts (e.g., displacement vs. distance).Essential Properties & Algebraic RulesMastery of these manipulations is required for "abstract function" problems where no explicit equation is given.Linearity: Additivity (Interval Addition): Reversing Limits: Zero Width: Dominance: If on , then .Niche Application (Geometric Evaluation): AP Exams often require evaluating integrals using geometry rather than antiderivatives. represents a semi-circle. represents a 'V' shape (triangles).Worked Example: Properties & GeometryGiven: and . Find: .1. Break down the target integral using Linearity2. Solve for using Additivity3. Evaluate the constant integral Method: Geometry (Rectangle with height 3, width ).4. Combine ResultsThe Fundamental Theorem of Calculus (Evaluation)Connection Between Derivatives & IntegralsOften called the Evaluation Theorem or FTC Part 1 in some texts, this theorem provides the practical method for evaluating definite integrals without Riemann sums. It connects differential and integral calculus: differentiation and integration are inverse operations.Requirements & NotationContinuity: must be continuous on .Antiderivative (): You must be able to find a function such that .Notation: The bracket notation is standard: .Common Pitfalls:Forgetting the Chain Rule (Reverse): When integrating , students often write instead of .Discontinuities: You cannot apply FTC directly across a vertical asymptote (e.g., diverges; blindly applying power rule yields a negative result, which is impossible for a positive function).Worked Example: Definite EvaluationEvaluate: 1. Find the AntiderivativeRecall Power Rule: Recall Log Rule: 2. Apply FTC ()3. SimplifyNote: The Second Fundamental Theorem (Accumulation)Derivative of an Accumulation FunctionThis theorem handles functions defined as integrals. It states that if you integrate a function and then differentiate it, you return to the original function (with variable adjustments). This models rate-of-change problems where the quantity is defined by accumulated area.Chain Rule Extension (Crucial)If the upper limit is a function rather than just , the Chain Rule applies. This is the most common version seen in FRQs.Key Components:Bottom Limit Constant: The lower limit must be a constant. If it is a variable, use properties to split the integral.Variable Swap: The variable inside the integral is a "dummy variable." It gets replaced by the expression in the upper limit.Worked Example: FTC with Chain RuleProblem: Let . Find .1. Identify StructureUpper limit is a function .Integrand is .Formula: .2. Differentiate Upper Limit3. Substitute and MultiplyReplace with : Multiply by derivative: Final Answer:Average Value & MVT for IntegralsConcept: "Leveling Out" the AreaAverage Value: If you melted the area under the curve into a perfect rectangle with width , what would the height be? That height is the average value.Mean Value Theorem (Integrals): Guarantees that a continuous function actually attains its average value at least once in the interval.Formulas & GeometryAverage Value Formula:MVT for Integrals Statement: If is continuous on , there exists at least one number in such that:Or equivalently: .Nuance: Distinguish between Average Rate of Change (Slope of secant: ) and Average Value of a Function (Integral height: ).Worked Example: Finding 'c'Problem: Find the value guaranteed by the MVT for Integrals for on .1. Calculate the Integral2. Calculate Average Value3. Set and Solve4. Check Interval Discard (outside interval).Keep .Rectilinear Motion (Integration Applications)From Acceleration to PositionIntegration allows us to move "up" the ladder of motion:where is position, is velocity, and is acceleration.Displacement vs. Total Distance (High Priority)This is the single most common trap in particle motion problems.Displacement (Net Change): How far you are from where you started.Total Distance Traveled: The sum of all ground covered (forward and backward). Requires absolute value.Current Position Formula: To find the actual position given an initial position :Worked Example: Motion AnalysisGiven: A particle moves with velocity m/s. Initial position . Find:Displacement on .Total Distance Traveled on .Position at .1. Displacement2. Total Distance (Requires splitting at zeros)Set : .Function is negative on and positive on .Split integral and apply absolute value to negative regions:Evaluate Part 1 ( to ):Evaluate Part 2 ( to ):Total Distance: .3. Position at Area Under a Curve & Riemann SumsApproximation & Limit DefinitionRiemann Sums serve as the foundational method for approximating the area bounded by a function curve and the x-axis. By partitioning an interval into sub-intervals and constructing geometric shapes (rectangles or trapezoids), we approximate the total area.This concept bridges pre-calculus geometry and calculus. The Definite Integral is rigorously defined as the limit of a Riemann Sum as the number of sub-intervals () approaches infinity (or as the width of the sub-intervals, , approaches zero).Summation Methods, Error Analysis, & NuancesThe approximation method depends on where the height of the rectangle is sampled within the sub-interval.LRAM (Left Riemann Sum): Height is at the left endpoint.Nuance: Underestimates if is increasing. Overestimates if is decreasing.RRAM (Right Riemann Sum): Height is at the right endpoint.Nuance: Overestimates if is increasing. Underestimates if is decreasing.MRAM (Midpoint Riemann Sum): Height is at the midpoint.Accuracy: Generally more accurate than LRAM or RRAM but requires calculating midpoint coordinates.Trapezoidal Rule: Uses trapezoids connecting points and .Formula: .Error Analysis (Concavity): The error depends on concavity, not slope.Concave Up: Overestimates (secant lines lie above curve).Concave Down: Underestimates (secant lines lie below curve).Common Pitfalls:Assuming RRAM always overestimates (it depends on slope, not concavity).Confusing "Total Area" (geometric area) with "Net Area" (integral value). Riemann sums calculate net signed area.Unequal sub-intervals: Standard formulas assume equal , but AP table problems often use unequal widths. You must calculate the area of each shape individually.Worked Examples: LRAM and Trapezoidal RuleScenario: Estimate the area under on using 3 equal sub-intervals.1. Setup Interval Width ()Intervals: 2. LRAM Calculation (Left Endpoints: 0, 2, 4)Calculate heights:Sum areas (Width Height):Analysis: Since is increasing, LRAM () is an underestimate.3. Trapezoidal ApproximationUse endpoints: .Apply formula :Analysis: Since is concave up (), Trapezoidal () is an overestimate.Definite Integrals & PropertiesSigned Area & Formal DefinitionA Definite Integral, denoted , represents the net signed area bounded by the curve and the x-axis from to .Area above the x-axis is positive.Area below the x-axis is negative.This distinction is vital for interpreting physical contexts (e.g., displacement vs. distance).Essential Properties & Algebraic RulesMastery of these manipulations is required for "abstract function" problems where no explicit equation is given.Linearity: Additivity (Interval Addition): Reversing Limits: Zero Width: Dominance: If on , then .Niche Application (Geometric Evaluation): AP Exams often require evaluating integrals using geometry rather than antiderivatives. represents a semi-circle. represents a 'V' shape (triangles).Worked Example: Properties & GeometryGiven: and . Find: .1. Break down the target integral using Linearity2. Solve for using Additivity3. Evaluate the constant integral Method: Geometry (Rectangle with height 3, width ).4. Combine ResultsThe Fundamental Theorem of Calculus (Evaluation)Connection Between Derivatives & IntegralsOften called the Evaluation Theorem or FTC Part 1 in some texts, this theorem provides the practical method for evaluating definite integrals without Riemann sums. It connects differential and integral calculus: differentiation and integration are inverse operations.Requirements & NotationContinuity: must be continuous on .Antiderivative (): You must be able to find a function such that .Notation: The bracket notation is standard: .Common Pitfalls:Forgetting the Chain Rule (Reverse): When integrating , students often write instead of .Discontinuities: You cannot apply FTC directly across a vertical asymptote (e.g., diverges; blindly applying power rule yields a negative result, which is impossible for a positive function).Worked Example: Definite EvaluationEvaluate: 1. Find the AntiderivativeRecall Power Rule: Recall Log Rule: 2. Apply FTC ()3. SimplifyNote: The Second Fundamental Theorem (Accumulation)Derivative of an Accumulation FunctionThis theorem handles functions defined as integrals. It states that if you integrate a function and then differentiate it, you return to the original function (with variable adjustments). This models rate-of-change problems where the quantity is defined by accumulated area.Chain Rule Extension (Crucial)If the upper limit is a function rather than just , the Chain Rule applies. This is the most common version seen in FRQs.Key Components:Bottom Limit Constant: The lower limit must be a constant. If it is a variable, use properties to split the integral.Variable Swap: The variable inside the integral is a "dummy variable." It gets replaced by the expression in the upper limit.Worked Example: FTC with Chain RuleProblem: Let . Find .1. Identify StructureUpper limit is a function .Integrand is .Formula: .2. Differentiate Upper Limit3. Substitute and MultiplyReplace with : Multiply by derivative: Final Answer:Average Value & MVT for IntegralsConcept: "Leveling Out" the AreaAverage Value: If you melted the area under the curve into a perfect rectangle with width , what would the height be? That height is the average value.Mean Value Theorem (Integrals): Guarantees that a continuous function actually attains its average value at least once in the interval.Formulas & GeometryAverage Value Formula:MVT for Integrals Statement: If is continuous on , there exists at least one number in such that:Or equivalently: .Nuance: Distinguish between Average Rate of Change (Slope of secant: ) and Average Value of a Function (Integral height: ).Worked Example: Finding 'c'Problem: Find the value guaranteed by the MVT for Integrals for on .1. Calculate the Integral2. Calculate Average Value3. Set and Solve4. Check Interval Discard (outside interval).Keep .Rectilinear Motion (Integration Applications)From Acceleration to PositionIntegration allows us to move "up" the ladder of motion:where is position, is velocity, and is acceleration.Displacement vs. Total Distance (High Priority)This is the single most common trap in particle motion problems.Displacement (Net Change): How far you are from where you started.Total Distance Traveled: The sum of all ground covered (forward and backward). Requires absolute value.Current Position Formula: To find the actual position given an initial position :Worked Example: Motion AnalysisGiven: A particle moves with velocity m/s. Initial position . Find:Displacement on .Total Distance Traveled on .Position at .1. Displacement2. Total Distance (Requires splitting at zeros)Set : .Function is negative on and positive on .Split integral and apply absolute value to negative regions:Evaluate Part 1 ( to ):Evaluate Part 2 ( to ):Total Distance: .3. Position at