top of page

Check the analytical derivative or integral on the Ti-84

Checking the analytically derivative  is very easy

Check your analytical derivatives or integrals with the TI-84 Plus CE. It is very useful for students in electrical and mechatronics engineering.

With some examples, we will show you how you can check your answers! Let us start by checking the derivate of a function.

Example 1




If you have found the analytical solution and want to check if it is the correct answer, the steps are shown in the following figures.

The function to be analyzed
The analytically solution
Example how it works to check your solution

Step 1: Enter the math problem into Y1.
Step 2: Enter the analytical solution you have derived into Y2.
Step 3: Calculate the difference between the TI-84 solution and your solution by entering Y3=Y1-Y2.
Step 4 : Choose the scales for Xmax and Xmin (the area where you want to compare the solutions). Also choose Ymax and Ymin or use the command "auto zoom".
Step 5 : Use the graph command to view the graphic solutions
The difference between Ti84 solution and the analytical solution : Y3 =Y1-Y2 =0 for the whole area fro Xmin to Xmax.Therefore you may conclude that the analytical solution is correct.
In the same way you can check integrals, differential equations etc., we will give more examples.

Check your analytically derived integral with the results of the integral calculated by your Ti-84.

In the same way, you can check your analytical solution of an integral. Suppose next integral with the solution I have derived. The integral which has to be calculated by the Ti84 is substituted in Y1. For the lower limit we choose 0 and the upper limit we choose X instead of a fixed value. Your proposed solution in substituted in Y2. The input for the Ti84 is shown. For the Window range Xmin-Xmax we choose 0-π.

Example of an integral exercise to be checked with wrong solution
Ti-84 used for checking the analytically solution  which seems wrong

The results show (step5) that our derived solution is not correct because Y3 is not equal to zero for the chosen X-range. The error we have made is that we have forgotten the minus sign;  thus the correct solution is :

Ti-84 used for checking the analytically solution  which is rightl

We found the correct solution because Y1=Y2 and Y3=0.
If an indefinite integral is asked the answer = : Y(x)-Y(0)=F(x)= -(1/2)cos(2x)+1/2 and therefore Y(x)=-(1/2)cos(2x)+C with C=Y(0)+1/2. The explained method is a good way to check your solution of the analytically derived integral. 

bottom of page