The problem tells us to approximate the integral using the midpoint rule. n=5, sin(x)2 dx 0 to 1. I understand that the answer is .2(sin(.1)^2 + sin(.3)^...). When I use my calculator to get the decimal answer, I cannot get the same answer the book has. I don't know what I am doing wrong.

To approximate the integral using the midpoint rule, you need to follow these steps:

1. Divide the interval [0, 1] into n subintervals of equal width. In this case, n = 5, so you'll have 5 subintervals.

2. Calculate the midpoint of each subinterval by adding the left endpoint and the right endpoint of each subinterval and dividing by 2. For example, for the first subinterval [0, 1/5], the midpoint is (0 + 1/5) / 2 = 1/10. Similarly, for the second subinterval [1/5, 2/5], the midpoint is (1/5 + 2/5) / 2 = 3/10.

3. Evaluate the function sin(x)^2 at each midpoint. For example, for the first subinterval, sin(1/10)^2. For the second subinterval, sin(3/10)^2, and so on.

4. Sum up all the evaluated values from step 3.

5. Multiply the sum from step 4 by the width of each subinterval, which is 1/n. In this case, the width is 1/5.

Let's go through the calculations:

For the first subinterval: sin(1/10)^2 = 0.00999983333416666^2 ≈ 0.000099999166707769.
For the second subinterval: sin(3/10)^2 = 0.29552020666134^2 ≈ 0.0873333333333333.
For the third subinterval: sin(5/10)^2 = sin(1/2)^2 = 1^2 = 1.
For the fourth subinterval: sin(7/10)^2 = 0.634963914784736^2 ≈ 0.402444682098765.
For the fifth subinterval: sin(9/10)^2 ≈ 0.000099999166707769.

Now, sum up all the values: 0.000099999166707769 + 0.0873333333333333 + 1 + 0.402444682098765 + 0.000099999166707769 = 1.49087718076861.

Finally, multiply the sum by the width of each subinterval: 1.49087718076861 * (1/5) = 0.298175436153723.

Therefore, the approximate value of the integral using the midpoint rule for n = 5 is approximately 0.298175436153723.

If your calculator is not giving you the same result, double-check that you are following the steps correctly and entering the calculations accurately. Note that calculators may have different rounding methods, so your result might vary slightly from the one given in the book.