Calculus

Furthermore

Official Guidance, clarification and syllabus links:

$$\int_{a}^{b} g'(x) \; dx = g(b)-g(a)$$

The value of some definite integrals can only be found using technology.

Link to: definite integrals using technology

Students are expected to first write a correct expression before calculating the area.

Technology may be used to enhance understanding of the relationship between integrals and areas.

Formula Booklet:

Area of region enclosed by a curve and x-axis

$$A = \int_{a}^{b} |y| \; dx $$

Areas between curves

We want to find the area between the curves \( y = x^2 + x - 2 \) and \( y = x + 2 \). First, we need to find the points of intersection between the two curves.

We can do this by setting the two equations equal to each other and solving for \( x \):

\[ \begin{align*} x^2 + x - 2 &= x + 2 \\ x^2 &= 4 \\ (x - 2)(x + 2) &= 0. \end{align*} \]

The solutions are \( x = 2 \) and \( x = -2 \), so the curves intersect at these points.

Next, we'll find the area between the curves by integrating the difference between the two functions over the interval from \(-2\) to \(2\)

\[ \begin{align*} \text{Area} &= \int_{-2}^{2} \left( x + 2 - (x^2 + x - 2) \right) \,dx \\ &= \int_{-2}^{2} \left( -x^2 + 4 \right) \,dx \\ &= \left[ - \frac{x^3}{3} + 4x \right]_{-2}^2\\ &=(-8/3+8)-(8/3-8)\\ &=10 \frac23 \text{ square units} \end{align*} \]

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

Trapezium Rule is a textbook style page with a few worked examples from Alevelmaths.co.uk.