# How to set Matlab do display all calculation results?

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Daniyar on 6 May 2013
Answered: Emmanuel on 19 May 2024
Hi,
How can I set Matlab to display all the calculation results?
I am solving the system of ODEs with large spatial variable and time step distribution. If I set total time to 100 with time step of 1 only the results starting from t=50 are displayed.
How to fix it?
Thanks

Babak on 6 May 2013
When the plotted data does not correspond to the whole "time span", it is usually because of that the time span is too large in comparison to the time step. So the ratio of timespan to timestep is large. Reducing it will show you more in the scope.
What do you mean by "spatial variable"? Are you solving a PDE that you have space variable? an ODE only has temporal variable and no spatial is involved.

Maikel on 8 May 2013
Go to preferences->Command Window->Number of lines in command window
And increase the value

Emmanuel on 19 May 2024
Write a MATLAB function named areaIntegration that takes multiple input arguments representing different shapes and their coordinates. The function should integrate the areas made up of these shapes using Boolean algebra and return the total integrated area. The function signature should be: function totalArea = areaIntegration(shape1, shape2,...) Where shape1, shape2, etc. are arrays representing different shapes, their scale, operation, and location. Each cell array should contain 5 elements such that: • 1 st Element is the shape identifier where “1” represent a square and “0” present a circle. • 2 nd Element is the scale of the shape. • 3 rd Element is the type of operation where ‘0’ is Union, ‘1’ is Subtract, and ‘2’ is Intersection with the shape of the previous argument. • 4 th and 5th is the location of the shape in Cartesian coordinate (i.e. x and y). Together they form an array like this: [shape identifier, scale, operation type, x-coordinate, y-coordinate]. For example: 1. If the 1st argument is an array [1, 1, 2, 0, 0], it means it is a square with a 1 unit dimension located at (0, 0), intersecting with the existing 2D space. 2. The 2nd argument, an array with the value [0, 2, 0, -1, 1], represents a circle with a 2 unit radius located at (-1, 1) to union with the result left by the 1st argument.

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