www.mathsrevision.com Higher Higher Unit 1 www.mathsrevision.com What is Integration The Process of Integration Area between to curves Application 1.4.

Slides:



Advertisements
Similar presentations
Higher Maths Integration
Advertisements

Higher Unit 2 Outcome 2 What is Integration
Integration Higher Mathematics Next.
Higher Unit 3 Differentiation The Chain Rule
“Teach A Level Maths” Vol. 1: AS Core Modules
Warm Up.
Follow the link to the slide. Then click on the figure to play the animation. A Figure Figure
Straight Line (2) Composite Functions (1) Higher Past Papers by Topic 2015 Onwards Higher Past Papers by Topic 2015 Onwards Differentiation.
6.5 The Definite Integral In our definition of net signed area, we assumed that for each positive number n, the Interval [a, b] was subdivided into n subintervals.
7 INVERSE FUNCTIONS.
Applications Of The Definite Integral The Area under the curve of a function The area between two curves The Volume of the Solid of revolution.
Straight Line (11) Composite Functions (12) Higher Past Papers by Topic Higher Past Papers by Topic Differentiation.
Do Now: p.381: #8 Integrate the two parts separately: Shaded Area =
Area Between Two Curves
Antiderivatives and the Rules of Integration
Higher Maths Question Types. Functions & Graphs TYPE questions (Trig, Quadratics) Sketching Graphs Composite Functions Steps : 1.Outside function stays.
Differentiation Higher Mathematics Next.
Area Between Two Curves
Properties of Logarithms
Integration. Indefinite Integral Suppose we know that a graph has gradient –2, what is the equation of the graph? There are many possible equations for.
Higher Unit 1 Finding the gradient for a polynomial Differentiating a polynomial Differentiating Negative.
Higher Unit 3 Further Differentiation Trig Functions Further Integration Integrating Trig Functions Differentiation The Chain Rule.
Differential calculation of function many the variables. Derivate of part. Part and complete differentials. Application complete a differential.
5.5 The Substitution Rule In this section, we will learn: To substitute a new variable in place of an existing expression in a function, making integration.
The Integral chapter 5 The Indefinite Integral Substitution The Definite Integral As a Sum The Definite Integral As Area The Definite Integral: The Fundamental.
Higher Unit 2 EF Higher Unit 3 Vectors and Scalars Properties of vectors Adding / Sub of vectors Multiplication.
Introduction This chapter gives you several methods which can be used to solve complicated equations to given levels of accuracy These are similar to.
7 Applications of Integration
CALCULUS II Chapter 5.
Section 7.2a Area between curves.
Higher Unit 1 Applications Finding the gradient for a polynomial Differentiating Easy Functions Differentiating.
Warmup 1) 2). 5.4: Fundamental Theorem of Calculus.
P1 Chapter 16 CIE Centre A-level Pure Maths © Adam Gibson.
AP CALCULUS AB PRACTICE EXAM. 1)Multiply by clever form of 1 3 and 1/3.
Introduction This chapter focuses on Parametric equations Parametric equations split a ‘Cartesian’ equation into an x and y ‘component’ They are used.
Chapter 6 The Definite Integral.  Antidifferentiation  Areas and Riemann Sums  Definite Integrals and the Fundamental Theorem  Areas in the xy-Plane.
Higher Expressions & Formulae Higher Unit 2 Exponential & Log Graphs Special “e” and Links between Log and.
Chapter 5: The Definite Integral Section 5.2: Definite Integrals
Section 5.4 Theorems About Definite Integrals. Properties of Limits of Integration If a, b, and c are any numbers and f is a continuous function, then.
INTEGRALS The Substitution Rule In this section, we will learn: To substitute a new variable in place of an existing expression in a function,
Introduction We have seen how to Integrate in C1 In C2 we start to use Integration, to work out areas below curves It is increasingly important in this.
Maths revision course by Miriam Hanks
P1 Chapter 12 & 15 CIE Centre A-level Pure Maths © Adam Gibson.
Application: Area under and between Curve(s) Volume Generated
Introduction Integration is the reverse process of Differentiation Differentiating gives us a formula for the gradient Integrating can get us the formula.
Higher Higher Unit 2 What is Integration The Process of Integration Integration & Area of a Curve Exam Type.
Higher Outcome 2 Higher Unit 3 Further Differentiation Trig Functions Harder Type Questions Further Integration.
Definite Integration and Areas 01 It can be used to find an area bounded, in part, by a curve e.g. gives the area shaded on the graph The limits of integration...
Integration – Overall Objectives  Integration as the inverse of differentiation  Definite and indefinite integrals  Area under the curve.
Chapter 6 Integration Section 5 The Fundamental Theorem of Calculus (Day 1)
5.3 Definite Integrals. Example: Find the area under the curve from x = 1 to x = 2. The best we can do as of now is approximate with rectangles.
Integration Part 1 Anti-Differentiation Integration can be thought of as the opposite of differentiation (just as subtraction is the opposite of addition).
Definite Integrals & Riemann Sums
Announcements Topics: -sections 7.3 (definite integrals) and 7.4 (FTC) * Read these sections and study solved examples in your textbook! Work On: -Practice.
Section 4.2 The Definite Integral. If f is a continuous function defined for a ≤ x ≤ b, we divide the interval [a, b] into n subintervals of equal width.
Definite Integrals, The Fundamental Theorem of Calculus Parts 1 and 2 And the Mean Value Theorem for Integrals.
Announcements Topics: -sections 7.3 (definite integrals) and 7.4 (FTC) * Read these sections and study solved examples in your textbook! Work On: -Practice.
Chapter 6 The Definite Integral.  Antidifferentiation  Areas  Definite Integrals and the Fundamental Theorem  Areas in the xy-Plane  Applications.
Announcements Topics: -sections 7.3 (definite integrals), 7.4 (FTC), and 7.5 (additional techniques of integration) * Read these sections and study solved.
Essential Question: How is a definite integral related to area ?
The Fundamental Theorem of Calculus Area and The Definite Integral OBJECTIVES  Evaluate a definite integral.  Find the area under a curve over a given.
Y =dy/dx 2x x2x2 3x 2 x3x3 x 2 + 2x 8x x 3 + 3x 2 + x 4 5x 4 y =dy/dx x x x x x 3 + x 2 + 2x 5x 4 + x x -2 + x.
Area of a Region Between 2 Curves
Warm Up.
Higher Unit 1 Applications 1.2
Coordinate Geometry in the (x,y) plane.
Chapter 6 The Definite Integral
Integration Higher Mathematics Next.
Tutorial 6 The Definite Integral
Using differentiation (Application)
Presentation transcript:

Higher Higher Unit 1 What is Integration The Process of Integration Area between to curves Application 1.4 Calculus Area under a curve Working backwards to find function Area under a curve above and below x-axis Exam

Higher Application 1.4 Calculus Integration we get You have 1 minute to come up with the rule. Integration can be thought of as the opposite of differentiation (just as subtraction is the opposite of addition).

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 Application 1.4 Calculus

Higher Integrating is the opposite of differentiating, so: integrate But: differentiate integrate Integrating 6x … which function do we get back to? Integration Application 1.4 Calculus

Higher Solution: When you integrate a function remember to add the Constant of Integration …………… + C Integration Application 1.4 Calculus

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 Application 1.4 Calculus

Higher Examples: Integration Application 1.4 Calculus

Higher Integration Application 1.4 Calculus Just like differentiation, we must arrange the function as a series of powers of x before we integrate.

Name : Integration techniques Area under curve = Area under curve = Integration

Higher Extra Practice Application 1.4 Calculus HHM Ex9G and Ex9H HHMEx9I Q1 a,b,e,fi,j,m,n,q,r

Definite Integrals Evaluate

Definite Integrals Evaluate

Definite Integrals Evaluate

Definite Integrals Find p, given

Higher Extra Practice Application 1.4 Calculus HHM Ex9K and Ex9L Q1, Q2

Real Application of Integration Find area between the function and the x-axis between x = 0 and x = 5 A = ½ bh = ½ x 5 x 5 = 12.5

Real Application of Integration Find area between the function and the x-axis between x = 0 and x = 4 A = ½ bh = ½ x 4 x 4 = 8 A = lb = 4 x 4 = 16 A T = = 24

Real Application of Integration Find area between the function and the x-axis between x = 0 and x = 2

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 Application 1.4 Calculus

Real Application of Integration Find area between the function and the x-axis between x = -3 and x = 3 ? Houston we have a problem !

We need to do separate integrations for above and below the x-axis. Real Application of Integration Areas under the x-axis ALWAYS give negative values By convention we simply take the positive value since we cannot get a negative area.

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 Application 1.4 Calculus

Integrate the function g(x) = x(x - 4) between x = 0 to x = 5 Real Application of Integration We need to sketch the function and find the roots before we can integrate

We need to do separate integrations for above and below the x-axis. Real Application of Integration Since under x-axis take positive value

Higher Extra Practice Application 1.4 Calculus HHM Ex9M and Ex9N

Higher The Area Between Two Curves To find the area between two curves we evaluate: Area under a Curve Application 1.4 Calculus

Find upper and lower limits. Area between Two Functions then integrate top curve – bottom curve.

Find upper and lower limits. then integrate top curve – bottom curve. Take out common factor

Higher Extra Practice Application 1.4 Calculus HHM Ex9K and Ex9L Q1, Q2

Higher To get the function f(x) from the derivative f’(x) we do the opposite, i.e. we integrate. Hence: Integration Application 1.4 Calculus

Higher Integration Application 1.4 Calculus Example :

Higher Extra Practice Application 1.4 Calculus HHM Ex9Q

Calculus Revision Back Next Quit Integrate Integrate term by term simplif y

Calculus Revision Back Next Quit Integrate Integrate term by term

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Split into separate fractions

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Multiply out brackets Integrate term by term simplify

Calculus Revision Back Next Quit Integrate Standard Integral (from Chain Rule)

Calculus Revision Back Next Quit Integrate Split into separate fractions Multiply out brackets

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

Calculus Revision Back Next Quit A curve for which passes through the point (–1, 2). Express y in terms of x. Use the point

Higher Examples: Area under a Curve Application 1.4 Calculus

Higher Example: Area under a Curve Application 1.4 Calculus

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 Application 1.4 Calculus 9 1

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

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 Application 1.4 Calculus

Higher Exam Type Questions Application 1.4 Calculus At this stage in the course we can only do Polynomial integration questions. In Unit 3 we will tackle trigonometry integration

Higher Application 1.4 Calculus Are you on Target ! Update you log book Make sure you complete and correct ALL of the Integration questions inIntegration the past paper booklet.