\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Calculus

Syllabus Content

Introduction to integration as anti-differentiation of functions of the form f(x)=axn+bxn-1+.... Anti-differentiation with a boundary condition to determine the constant term. Definite integrals using technology. Area of a region enclosed by a curve y=f(x) and the x-axis, where f(x)>0

Here are some specific activities, investigations or visual aids we have picked out. Click anywhere in the grey area to access the resource.

Here are some exam-style questions on this statement:

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Click on a topic below for suggested lesson Starters, resources and activities from Transum.


Furthermore

Official Guidance, clarification and syllabus links:

Students should be aware of the link between anti-derivatives, definite integrals and area.

Example: if \( \frac{dy}{dx}=3x^2 + x \) and \(y=10\) when \(x=1\), then \(y=x^3+\frac12 x^2 + 8.5\)

Students are expected to first write a correct expression before calculating the area, for example

$$ \int^6_2(3x^2+4)dx $$

The use of dynamic geometry or graphing software is encouraged in the development of this concept.


Formula Booklet:

Integral of \(x^n\)

\( \int x^n dx = \dfrac{x^{n+1}}{n+1} + C, n \neq -1\)

Area between a curve \(y=f(x)\) and the x-axis, where \( f(x) \gt 0 \)

\( A = \int^b_a y dx\)


Integration is a fundamental concept in calculus that represents the accumulation of quantities. Specifically, when considering the area of a region enclosed by a curve \( y = f(x) \) and the x-axis, where \( f(x) > 0 \), integration allows us to calculate this area. The area under the curve from \( x = a \) to \( x = b \) is given by the definite integral:

$$ \int_a^b f(x) \, dx $$

For example, let's consider the function \( f(x) = x^2 \) over the interval [0, 2]. The area under the curve and above the x-axis between these two points can be found using the integral:

$$ \int_0^2 x^2 \, dx \\ = \left. \frac{1}{3}x^3 \right|_0^2 \\ = \frac{1}{3}(2^3) - \frac{1}{3}(0^3) \\ = \frac{8}{3} $$

Thus, the area of the region enclosed by the curve \( y = x^2 \) and the x-axis between \( x = 0 \) and \( x = 2 \) is \( \frac{8}{3} \) square units.


If you use a TI-Nspire GDC there are instructions useful for this topic.


This video on Antidifferentiation is from Revision Village and is aimed at students taking the IB Maths Standard level course


This video titled Overview of Integral Calculus is from Revision Village and is aimed at students taking the IB Maths Standard level course.


This Finding Areas Under Curves video is from Revision Village and is aimed at students taking the IB Maths Standard level course


How do you teach this topic? Do you have any tips or suggestions for other teachers? It is always useful to receive feedback and helps make these free resources even more useful for Maths teachers anywhere in the world. Click here to enter your comments.


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