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www.mathsrevision.com Higher Higher Unit 2 www.mathsrevision.com What is Integration The Process of Integration Integration & Area of a Curve Exam Type Questions Higher Outcome 2
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www.mathsrevision.com Higher Integration Part 1 Anti-Differentiation Integration can be thought of as the opposite of differentiation (just as subtraction is the opposite of addition). In general:DifferentiatingIntegrating Confusing? Is there any easier way? Outcome 2
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www.mathsrevision.com Higher Differentiation multiply by power decrease power by 1 Integration increase power by 1 divide by new power Where does this + C come from? Integration Outcome 2
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www.mathsrevision.com Higher Integrating is the opposite of differentiating, so: integrate But: differentiate integrate Integrating 6x ….......which function do we get back to? Integration Outcome 2
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www.mathsrevision.com Higher Solution: When you integrate a function remember to add the Constant of Integration …………… + C Integration Outcome 2
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www.mathsrevision.com Higher means “integrate 6x with respect to x” means “integrate f(x) with respect to x” Notation This notation was “invented” by Gottfried Wilhelm von Leibniz Integration Outcome 2
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www.mathsrevision.com Higher Examples: Integration Outcome 2
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www.mathsrevision.com Higher Integration Outcome 2 Just like differentiation, we must arrange the function as a series of powers of x before we integrate.
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www.mathsrevision.com Higher To get the function F(x) from the derivative F’(x) we do the opposite, i.e. we integrate. Hence : Integration Outcome 2
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www.mathsrevision.com Higher Further examples of integration Exam Standard Integration Outcome 2
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www.mathsrevision.com Higher The integral of a function can be used to determine the area between the x-axis and the graph of the function. NB:this is a definite integral. It has lower limit a and an upper limit b. Area under a Curve Outcome 2
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www.mathsrevision.com Higher Examples: Area under a Curve Outcome 2
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www.mathsrevision.com Higher Conventionally, the lower limit of a definite integral is always less then its upper limit. Area under a Curve Outcome 2
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www.mathsrevision.com Higher a b cd y=f(x) Very Important Note: When calculating integrals: areas above the x-axis are positive areas below the x-axis are negative When calculating the area between a curve and the x-axis: make a sketch calculate areas above and below the x-axis separately ignore the negative signs and add Area under a Curve Outcome 2
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www.mathsrevision.com Higher The Area Between Two Curves To find the area between two curves we evaluate: Area under a Curve Outcome 2
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www.mathsrevision.com Higher Example: Area under a Curve Outcome 2
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www.mathsrevision.com Higher Complicated Example: The cargo space of a small bulk carrier is 60m long. The shaded part of the diagram represents the uniform cross-section of this space. Find the area of this cross- section and hence find the volume of cargo that this ship can carry. Area under a Curve Outcome 2 9 1
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www.mathsrevision.com Higher The shape is symmetrical about the y-axis. So we calculate the area of one of the light shaded rectangles and one of the dark shaded wings. The area is then double their sum. The rectangle: let its width be s The wing: extends from x = s to x = t The area of a wing (W ) is given by: Area under a Curve
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www.mathsrevision.com Higher The area of a rectangle is given by: The area of the complete shaded area is given by: The cargo volume is: Area under a Curve Outcome 2
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www.mathsrevision.com Higher Exam Type Questions Outcome 2 At this stage in the course we can only do Polynomial integration questions. In Unit 3 we will tackle trigonometry integration
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Integration Higher Mathematics www.maths4scotland.co.uk Next
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Calculus Revision Back Next Quit Integrate Integrate term by term simplif y
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Calculus Revision Back Next Quit Find
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Calculus Revision Back Next Quit Integrate Multiply out brackets Integrate term by term simplify
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Calculus Revision Back Next Quit Find
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Calculus Revision Back Next Quit Integrate Standard Integral (from Chain Rule)
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Calculus Revision Back Next Quit Find p, given
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Calculus Revision Back Next Quit Evaluate Straight line form
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Calculus Revision Back Next Quit Find Use standard Integral (from chain rule)
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Calculus Revision Back Next Quit Find Integrate term by term
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Calculus Revision Back Next Quit Integrate Straight line form
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Calculus Revision Back Next Quit Integrate Straight line form
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Calculus Revision Back Next Quit Integrate Straight line form
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Calculus Revision Back Next Quit Integrate Split into separate fractions
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Calculus Revision Back Next Quit Find Use standard Integral (from chain rule)
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Calculus Revision Back Next Quit Find
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Calculus Revision Back Next Quit Find
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Calculus Revision Back Next Quit Integrate Straight line form
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Calculus Revision Back Next Quit Given the acceleration a is: If it starts at rest, find an expression for the velocity v where Starts at rest, so v = 0, when t = 0
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Calculus Revision Back Next Quit A curve for which passes through the point Find y in terms of x. Use the point
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Calculus Revision Back Next Quit Integrate Split into separate fractions Multiply out brackets
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Calculus Revision Back Next Quit If passes through the point express y in terms of x. Use the point
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Calculus Revision Back Next Quit Integrate Straight line form
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Calculus Revision Back Next Quit The graph of passes through the point (1, 2). express y in terms of x. If simplify Use the point Evaluate c
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Calculus Revision Back Next Quit Integrate Straight line form
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Calculus Revision Back Next Quit A curve for which passes through the point (–1, 2). Express y in terms of x. Use the point
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Calculus Revision Back Next Quit Evaluate Cannot use standard integral So multiply out
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Calculus Revision Back Next Quit Evaluate Straight line form
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Calculus Revision Back Next Quit Evaluate Use standard Integral (from chain rule)
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Calculus Revision Back Next Quit The curve passes through the point Find f(x ) use the given point
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Calculus Revision Back Next Quit Integrate Integrate term by term
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Calculus Revision Back Next Quit Integrate Integrate term by term
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Calculus Revision Back Next Quit Evaluate
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Quit C P D www.maths4scotland.co.uk © CPD 2004
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