Lesson Notes By Weeks and Term v4 - SHS 3
APPLICATION OF CALCULUS
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APPLICATION OF CALCULUS
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Subject: Additional Mathematics
Class: SHS 3
Term: 1st Term
Week: 14
Grade code: 3.3.2.LI.2
Strand code: 3
Sub-strand code: 2
Content standard code: 3.3.2.CS.1
Indicator code: 3.3.2.LI.2
Theme: CALCULUS
Subtheme: APPLICATION OF CALCULUS
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Calculus, particularly integration, is not just about abstract symbols and rules. It is a powerful tool used to solve real-world problems that involve changing quantities. In this lesson, we will explore two major applications: calculating the exact area of irregular shapes bounded by curves and analysing the motion of objects. Understanding these applications is crucial for fields like engineering, where we design roads and structures; in business, for calculating total profit; and in science, for modelling real-world phenomena. For example, engineers working on the Tema Motorway expansion would use these principles to calculate the volume of earth to be moved.
This lesson is divided into two main parts: finding areas and analysing motion (kinematics). Part A: Finding the Area Between Curves Area Between a Curve and the x-axis
The definite integral gives us the *net signed area* between a function `f(x)` and the x-axis from `x = a` to `x = b`. Net Signed Area: `Area = ∫[a to b] f(x) dx` If `f(x)` is above the x-axis (`f(x) > 0`), the integral is positive, and this value is the actual area. If `f(x)` is below the x-axis (`f(x) 0`, the line `y = x + 2` is the upper curve and `y = x²` is the lower curve in this interval. Step 3: Set up and evaluate the integral. Area `A = ∫[-1 to 2] [(x + 2) - (x²)] dx` `A = ∫[-1 to 2] (-x² + x + 2) dx`
Integrate: `[-(x³/3) + (x²/2) + 2x]` from -1 to 2.
Evaluate at the upper limit (`x = 2`): `[-(2³/3) + (2²/2) + 2(2)] = [-8/3 + 4/2 + 4] = [-8/3 + 2 + 4] = [-8/3 + 6] = 10/3`