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    IB Math AASL
    /
    Integration
    /

    Skills

    Skill Checklist

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    Integration

    Skill Checklist

    Track your progress across all skills in your objective. Mark your confidence level and identify areas to focus on.

    23 Skills Available

    Track your progress:

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    Working on it

    Confident

    📖 = included in formula booklet • 🚫 = not in formula booklet

    Track your progress:

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    Working on it

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    📖 = included in formula booklet • 🚫 = not in formula booklet

    Definite Integrals, Areas, and Basic Anti-Derivatives

    11 skills
    Definite Integrals
    SL 5.5

    A definite integral is evaluated between a lower and upper bound.

    ∫ab​f(x)dx is the integral of f(x)dx from x=a to x=b

    We can solve a definite integral with

    ∫ab​f(x)dx=[F(x)]ab​=F(b)−F(a)🚫


    where F(x)=∫f(x)dx.

    Watch video explanation →
    Area under a curve
    SL 5.5

    The area between a curve f(x)>0 and the x-axis is given by

    A=∫ab​f(x)dx📖

    Powered by Desmos

    Watch video explanation →
    Area between curve and x-axis
    SL 5.11

    In general, the area enclosed between a curve and the x-axis is given by

    A=∫ab​∣f(x)∣dx📖

    since any region below the x-axis has f(x)<0, but area must always be positive.

    Powered by Desmos

    This can be done with technology, or by splitting the integral into parts - where f is positive and where f is negative:

    A=∫am​f(x)dx+∫mb​−f(x)dx🚫
    Watch video explanation →
    Area between curves
    SL 5.11

    The area enclosed between two curves is given by

    A=∫ab​∣f(x)−g(x)∣dx

    Powered by Desmos

    This can be done with technology, or by splitting the integral into multiple regions, each having either f(x)>g(x) or g(x)>f(x).

    Watch video explanation →
    Definite Integral Rules
    SL 5.11

    Integrals of the same function with adjacent domains can be merged:

    ∫ab​f(x)dx+∫bc​f(x)dx=∫ac​f(x)dx🚫

    Similarly, the domain of an integral can be split:

    ∫ab​f(x)dx=∫am​f(x)dx+∫mb​f(x)dx🚫

    for any a<m<b.

    Watch video explanation →
    Integration as reverse differentiation
    SL 5.5

    Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbol∫ and write:

    ∫f(x)dx=a function with a derivative off(x)

    By convention we denote this function F:

    ∫f(x)dx=F(x)🚫

    We can also write

    ∫dxdy​dx=y🚫

    Notice the dx under the integral. This tells us which variable we are integrating with respect to - in this case we are reversing dxd​.

    Watch video explanation →
    The Integration Constant
    SL 5.5

    Since the derivative of a constant is always zero, then if if F′(x)=f(x), then (F(x)+C)′=f(x).


    This means that when we integrate, we can add any constant to our result, since differentiating makes this constant irrelevant:

    ∫f(x)dx=F(x)+C🚫
    Watch video explanation →
    Anti-Derivative of xⁿ, n∈ℤ
    SL 5.5
    ∫xndx=n+1xn+1​+C,n∈Z,n=−1📖
    Watch video explanation →
    Integrals of sums and scalar products
    SL 5.5

    In the same way that constant multiples can pass through the derivative, they can pass through the integral:

    ∫af(x)dx=a∫f(x)dx🚫

    And in the same way that the derivative of a sum is the sum of the derivatives:

    ∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx🚫
    Watch video explanation →
    Boundary Conditions
    SL 5.5

    If we know the value of y or f(x) for a given x, we can determine C by plugging in x and y.

    Watch video explanation →
    Calculating Definite Integral with GDC
    SL 5.5

    Graphing calculators can be used to evaluate definite integrals.


    For example, on a TI-84, math > 9:fnInt(, which prompts you with ∫□□​(□)d□. Make sure the variable of your function matches the variable that you take the integral with respect to.

    Anti-Derivative Rules

    4 skills
    Anti-Derivative of xⁿ, n∈ℚ
    SL 5.6
    ∫xndx=n+1xn+1​+C,n=−1📖
    Watch video explanation →
    Anti-Derivative of sin and cos
    SL 5.6

    The integrals of sin and cos are

    ∫sinxdx ∫cosxdx​=−cosx+C📖 =sinx+C📖​
    Watch video explanation →
    Anti-Derivative of e^x
    SL 5.6
    ∫exdx=ex+C📖
    Watch video explanation →
    Anti-Derivative of 1/x
    SL 5.6
    ∫x1​dx=ln∣x∣+C📖
    Watch video explanation →

    Techniques of Integration

    3 skills
    Integrating f(ax+b)
    SL 5.10

    If F(x)=∫f(x)dx, then

    ∫f(ax+b)dx=a1​F(ax+b)🚫
    Watch video explanation →
    Integration by substitution
    SL 5.10

    Integrating a composition of functions f(g(x)) requires us to divide by g′(x), so it is easier to find the anti-derivative of anything of the form g′(x)f′(g(x)) by first dividing by g′(x).


    In symbols, we use the known fact

    ∫kg′(x)f′(g(x))dx=kf(g(x))+C🚫

    and let u=g(x), giving us

    ∫kg′(x)f′(g(x))dx=k∫f′(u)du,🚫

    an integral we can solve more easily:

    k∫f′(u)du=kf(u)+C.


    Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.

    Watch video explanation →
    Substitution and Integral Bounds
    SL 5.10

    When we make a substitution in a definite integral in the form

    ∫ab​kg′(x)f′(g(x))dx

    we need to remember that the bounds are from x=a to x=b:

    ∫ab​kg′(x)f′(g(x))dx  ​=k∫x=ax=b​f′(u)du =[kf(u)]x=ax=b​​


    We then have two choices:

    1. Plug x=a and x=b into u to find the bounds in terms of u.

    2. Plug u(x) back in and use the bounds a→b.

    Watch video explanation →

    Kinematics

    5 skills
    Displacement, velocity and acceleration
    SL 5.9

    Acceleration is the rate of change of velocity, which is the rate of change of displacement.


    v=dtds​🚫 ​
    a=dtdv​=dt2d2s​📖


    Hence, the integral of acceleration is velocity, and the integral of velocity is displacement.

    Watch video explanation →
    Displacement vs Distance
    SL 5.9

    Displacement s is the change in position between start and end time, whereas distance d is the total length of the path taken.


    Direction does not matter for distance, which is never negative, but displacement can be negative - usually indicating motion down or to the left.

    Powered by Desmos

    Watch video explanation →
    Change in Displacement from Velocity
    SL 5.9

    Change in displacement between t1​,t2​:

    ∫t1​t2​​v(t)dt📖
    Watch video explanation →
    Distance from Velocity
    SL 5.9

    The distance can be found from the velocity using the equation

    ∫t1​t2​​∣v(t)∣dt📖


    Example:

    Given v(t)=5−t m/s, find distance traveled between t=1 and t=10:

    d=∫19​∣5−t∣dt=∫15​(5−t)dt+∫59​(t−5)dt
    =[5t−2t2​]15−[5t−2t2​]59
    =(25−225​)−(5−21​)−(45−281​)+(25−225​)
    =25−5−45+25+2−25+1+1+25​
    =0+232​=16m
    Watch video explanation →
    Speed
    SL 5.9

    Speed is the magnitude of velocity:

    speed=∣v∣
    Watch video explanation →