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

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

- Integration Exercises on indefinite and definite integration of basic algebraic and trigonometric functions.
- Integration Video You can't possibly learn all about integration from a 28 minute video so all that this resource can do is provide a quick revision to help you do the online exercise.

Here are some exam-style questions on this statement:

- "
*Find:*" ... more - "
*(a) Find \(\int (4x+5) dx\).*" ... more - "
*A particle moves in a straight line. During the first nine seconds the velocity, \(v\) ms*^{-1}of the particle at time \(t\) seconds is given by:$$ v(t) = t \cos(t+5) $$

*The following diagram shows the graph of v:*" ... more - "
*Consider the region where \(0 \lt x \lt 2\pi \) and \(\sin{x} \gt \cos{2x} \)*" ... more - "
*The function \(f\) is defined by \(f(x) = 8 - 5 \sin{x} \), for \( x \ge 0 \).*" ... more - "
*A particle moves in a straight line such that its velocity, \(v\) ms*" ... more^{-1}, at time \(t\) seconds is given by: - "
*The diagram shows part of the graph of \(f(x)=Ae^{kx}+2\).*" ... more - "
*Consider the graph of the function \(f(x)=x^2+2\).*" ... more - "
*Make a sketch of a graph showing the velocity (in \(ms^{-1}\)) against time of a particle travelling for six seconds according to the equation:*" ... more - "
*Find the value of \(a\) if \(\pi \lt a \lt 2\pi\) and:*" ... more - "
*This graph represents the function \(f:x\to a \cos x, a\in \mathbf N\)*" ... more - "
*Let \(f(x) = \frac{ln3x}{kx} \) where \( x \gt 0\) and \( k \in \mathbf Q^+ \).*" ... more

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

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

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