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    IB Math AASL
    /
    Integration
    /

    Video

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    Watch comprehensive video reviews for most units, designed for final exam preparation. Each video includes integrated problems you can solve alongside detailed solutions.

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    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

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    Not your average video:

    Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

    Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

    Integration

    Video Reviews

    Watch comprehensive video reviews for Integration, designed for final exam preparation. Each video includes integrated problems you can solve alongside detailed solutions.

    Not your average video:

    Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

    Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

    Not your average video:

    Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

    Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

    Definite integralsKinematicsHL

    The video will automatically pause when it reaches a problem.

    Integration by substitution

    SL 5.10

    By noticing that dxd​[f(g(x))+C]=g′(x)⋅f′(g(x)) (chain rule), we can deduce that

    ∫kg′(x)f′(g(x))dx=kf(g(x))+C🚫

    This formula is quite obscure, but is made clear by substituting u=g(x)⇒g′(x)=dxdu​. The integral then becomes

    ∫kdxdu​⋅f′(u)dx=k∫f′(u)du=kf(u)+C🚫


    To find the correct u, we search for one part that is "almost" the derivative of another part. "Almost" because constant multiples k do not fundamentally alter the process.


    Example

    In the integral∫3x√x2−1​dx, we recognize that dxd​(x2−1)=2x which is a constant multiple of 3x. So we set u=x2−1, which gives dxdu​=2x. Thus the integral becomes:

    ∫23​⋅dxdu​⋅√udx    ​=23​∫u1/2du =u3/2+C =(x2−1)3/2+C​

    Integration by substitution

    SL 5.10

    By noticing that dxd​[f(g(x))+C]=g′(x)⋅f′(g(x)) (chain rule), we can deduce that

    ∫kg′(x)f′(g(x))dx=kf(g(x))+C🚫

    This formula is quite obscure, but is made clear by substituting u=g(x)⇒g′(x)=dxdu​. The integral then becomes

    ∫kdxdu​⋅f′(u)dx=k∫f′(u)du=kf(u)+C🚫


    To find the correct u, we search for one part that is "almost" the derivative of another part. "Almost" because constant multiples k do not fundamentally alter the process.


    Example

    In the integral∫3x√x2−1​dx, we recognize that dxd​(x2−1)=2x which is a constant multiple of 3x. So we set u=x2−1, which gives dxdu​=2x. Thus the integral becomes:

    ∫23​⋅dxdu​⋅√udx    ​=23​∫u1/2du =u3/2+C =(x2−1)3/2+C​
    Definite integralsKinematicsHL