Lesson 1

20.1.1

One antiderivative appears to be , as confirmed by the graph of Y2 = X4 / 4 coinciding with the graph of .

 [-5, 5, 1] x [-15, 15, 5] Y1 = fnInt(T^3,T,0,X) Y2 = X^4/4

Lesson 2

20.2.1

20.2.2

For x = 1 and y = 1,
Therefore, .

20.2.3

The graph shown is that of , the solution of with y(1) = 1 which was found earlier.

Lesson 3

20.3.1

You need 20 points to reach y(1) 1.95.

20.3.2

y(1.2) 1.2073. The actual value rounds to 1.2097.

Lesson 4

20.4.1

In this slope field the slopes of segments in a given row are equal. In the previous slope field the slopes of segments in a given column were equal. This difference is due to the fact that the differential equation in this problem is a function of y, therefore the slopes depend on y, not x.

Self Test

Y1 = fnInt(cos(T/2),T,0,X)

[-2 ,2 ,1] x [-2, 2, 1]

Any integral of the form is a member of the family of curves represented by .
The graph shown above is the graph for a = 0. Other values of a will produce vertical shifts of this graph.

[0, 1, 1] x [0, 15, 5]