Calculus - Integration - Quick Guides

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Study GuideCalculus–Integration1.Distance, Velocity, and AccelerationMany real-world calculus problems involvemotion, such as objects moving through space over time.In these problems,distance,velocity, andaccelerationare all functions of time, and calculus helpsus understand how they are connected.How These Ideas Are Related•Thederivative of distancegivesinstantaneous velocity.•Thederivative of velocitygivesinstantaneous acceleration.•Theindefinite integral of accelerationgives velocity.•Theindefinite integral of velocitygives distance.In short,derivatives describe how motion changes, andintegrals help us recover motion fromthose changes.Motion Under GravityFor objects moving straight up or down near Earth’s surface, gravity causes a constant accelerationof:The negative sign means the acceleration is directeddownward. As time increases, the velocitydecreases if the object is moving upward, or becomes more negative if the object is falling downward.Finding Velocity from AccelerationSince velocity is the integral of acceleration:

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Study GuideSubstitute(a(t) =-32):To find the constant(C1), use theinitial velocity.If the initial velocity at(t = 0)is(v0), then:So the velocity function becomes:Finding Distance from VelocityDistance is the integral of velocity:Substitute(v(t) =-32t + v0):To find(C2), use theinitial position(s0):So the distance function is:

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Study GuideExample1:Time to Hit the GroundA ball is thrown downward from a height of 512 feet with an initial velocity of 64 ft/sec. How long doesit take to reach the ground?Step 1: Identify the Given Information•(a(t) =-32)ft/sec²•(v0=-64)ft/sec (negative because it’s downward)•(s0= 512)ftStep 2: Write the Distance FunctionStep 3: Set Distance Equal to ZeroThe ball reaches the ground when(s(t) = 0):Factor:Step 4: SolveSince time cannot be negative:

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Study GuideFinal AnswerThe ball reaches the ground4 secondsafter it is thrown.Example2:Velocity When the Ball Hits the GroundWhat is the velocity of the ball when it hits the ground?Using the velocity function:Substitute(t = 4):Final AnswerThe ball hits the ground with a velocity of–192 ft/sec.The negative sign indicates the ball is movingdownward, meaning the distance from the ground isdecreasing as time increases.Example3:Missile Launch ProblemA missile accelerates upward at 4 m/sec² from rest in a silo 35 meters below ground level. How highabove the ground is it after 6 seconds?Step 1: Identify the Given Information•(a(t) = 4)m/sec²•(v0= 0)m/sec (starts from rest)•(s0=-35)m (below ground level)Step 2: Find the Velocity Function

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Study GuideStep 3: Find the Distance FunctionStep 4: Evaluate at(t = 6)Final AnswerAfter 6 seconds, the missile is109 meters above the ground.Key Takeaways•Velocity is the derivative of distance.•Acceleration is the derivative of velocity.•Integrals reverse derivatives and help build motion equations.•Initial conditions determine constants of integration.•Signs matter—they tell you the direction of motion.2.Definite IntegralsHow Definite Integrals Are DefinedTo understand the idea of a definite integral, we start with a function(f(x))that iscontinuouson aclosed interval ([a, b]).

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Study GuideStep 1: Divide the IntervalFirst, the interval ([a, b]) is divided into(n)smaller subintervals. These subintervals are often chosento have equal length, called (Δx), although this is not strictly required.Step 2: Choose Sample PointsFrom each subinterval, we select a point(xi). For each of these points, we evaluate the function to get(f(xi)).Step 3: Form ProductsNext, we multiply each function value by the width of its subinterval:Each of these products represents a small contribution to the total value we are trying to approximate.Step 4: Add Everything TogetherFinally, we add all these products to form aRiemann sum:or, using summation notation,Understanding the Riemann SumARiemann sumcan be thought of as a “sum of(n)products.” Its value depends on how the functionbehaves over the interval ([a, b]):•If(f(x) > 0)on ([a, b]), the Riemann sum ispositive.•If(f(x) < 0)on ([a, b]), the Riemann sum isnegative.•In some cases, positive and negative values may balance out, resulting in a sum ofzero.

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Study GuideRiemann sums are important because they provide the foundation for defining the definite integral. Aswe increase the number of subintervals and make them very small, the Riemann sum approaches theexact value of the definite integral.Example1:Approximating Area Using a Riemann SumLet’s work through an example to see how a Riemann sum is calculated step by step.Figure 1 A Riemann sum with four subintervals.ProblemEvaluate the Riemann sum forusingfour subintervals of equal length, where each(xi)is chosen as theright endpointof itssubinterval.

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Study GuideStep 1: Find the Width of Each SubintervalSince the interval ([1,3]) is divided into 4 equal parts, the width of each subinterval isStep 2: Identify the Right EndpointsStarting at(x = 1)and moving right by (Δx =1/2), the right endpoints are:Step 3: Set Up the Riemann SumThe Riemann sum for four subintervals is written as:Substituting the values:Step 4: Evaluate the Function ValuesSince(f(x) = x2):Add them:

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Study GuideNow multiply by (Δx =1/2):What Happens as We Use More Subintervals?As the number of subintervals increases:•Each subinterval becomesnarrower•The approximation becomesmore accurateMathematically, this means:•(n→∞)•(Δx→0)If the Riemann sum approaches a limit as this happens, that limit defines thedefinite integral.Definition of the Definite IntegralIf the limit exists, the definite integral of(f(x))on ([a,b]) is defined as:Important Terminology•Integrand: the function being integrated,(f(x))•Variable of integration: the variable(x)•Limits of integration:o(a): lower limito(b): upper limitEven though the symbol(∫)looks the same as in indefinite integrals, remember:•Adefinite integral gives a single real number•Anindefinite integral represents a family of functions

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Study GuideWhen Does a Definite Integral Exist?The existence of the limit of a Riemann sum is very important.•If a function(f(x))iscontinuouson a closed interval ([a,b]), then:oThe definite integralexistsoThe function is said to beintegrableon ([a,b])Continuity guarantees integrability, but the reverse is not always true.Also, even if a definite integral exists, it may not always be easy to calculate exactly.Properties of Definite IntegralsHere are some key rules that help us work with definite integrals:

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