Question

Use the midpoint rule with n=4 to approximate the region bounded by y=-x^3 and y=-x


Show that the function f(x)= (integral from 2x to 5x) (1)/(t) dt is constant on the interval (0,+infinity)

Answer 1

For the first one i think the integral is from -1 to 1 for the two functions, because they intersect at (-1,1) (0,0) and (1,-1)

Answer 2

(-1,-1), (0,0) and (1,1). The four midpoints are
-3/4, -1/4, +1/4, and +3/4. Each vertical strip has a width of 1/2. The lengths of the 4 strips are:
-27/64-(-3/4); -1/64-(-1/4); 1/4 - 1/64; and 3/4-27/64;
simplifying these, you get
21/64, 15/64, 15/64, and 21/64.
So the area is something like
(1/2)*(21+15+15+21)/64 = 36/64 = 9/16. MY ANSWER TO question 1

Answer 3

since -x^3 < -x on (-1,0), you should be subtracting (-x)-(-x^3). the actual math should be

-(-3/4) - (-(-3/4)^3) = 3/4 - 27/64 = 21/64
and so on. It comes out the same.

Assuming your intermediate results are good, the answer looks good.

Answer 4

Q1:
There are three intersection points: (-1,1), (0,0), (1,-1)
Thus the endpoints are: (-1,1), (-0.5,0.5), (0,0), (0.5,0.5), (1,1)
Thus, the midpoints are: (-0.75,0.75), (-0.125,0.125), (0.125,-0.125), (0.75,-0.75)
The area is as follows:
0.75(-0.5) + (0.25(-0.5)) + -0.25(0.5) + -0.75(0.5)
= -1
I took the area of the whole thing instead of taking the area of the 2 regions separately because I assumed that is what the question is asking
Q2:
If we evaluate the derivative of the function, we get:
int of [ln(t)] from 2x to 5x which results in
ln|5x| - ln|2x|
Taking the derivative of this, we get:
d/dx (ln(5|x|) - ln(2|x|))
d/dx (ln(5|x|/2|x|))
d/dx (ln(5/2))
This equals 0
Since the derivative of a function gives the slope of the tangent line at any given point, we know the above function is constant on the given interval because the derivative is 0 which indicates the function neither decreases nor increases which means it remains constant