Integration | Perplex

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

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Area between curves

SL 5.11

The area enclosed between two curves is given by

A=∫ab∣f(x)−g(x)∣dx

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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).

Example

Find the area enclosed between y=x3 and y=x.

The curves intersect when x3=x⇒x=0 or x2=1 ie x=±1.

For −1<x<0:

x3−x=x(x2−1)= negative× negative>0

For 0<x<1:

x3−x=x(x2−1)= positive× negative<0

So the area is

A =∫−11∣∣x3−x∣∣dx =∫−10x3−xdx+∫01−(x3−x)dx =[4x4−2x2]−10−[4x4−2x2]01 =[0−(41−21)]−[(41−21)−0]=21

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).

Example

Find the area enclosed between y=x3 and y=x.

The curves intersect when x3=x⇒x=0 or x2=1 ie x=±1.

For −1<x<0:

x3−x=x(x2−1)= negative× negative>0

For 0<x<1:

x3−x=x(x2−1)= positive× negative<0

So the area is

A =∫−11∣∣x3−x∣∣dx =∫−10x3−xdx+∫01−(x3−x)dx =[4x4−2x2]−10−[4x4−2x2]01 =[0−(41−21)]−[(41−21)−0]=21