How do you do problem #2 on section 24 in the homework?
The integral from 0 to 4 of f(t) = e^sqrt(t) using the t values 0, 2, and 3 and the left rectangular rule is done by calculating the (signed) area of the rectangles where the heights are calculated as f(t) where t is the left side of the rectangle. In our case there are three rectangles and so the sum is: f(0)(2 - 0) + f(2)(3 - 2) + f(3)(4 - 3) = 2*e^sqrt(0) + e^sqrt(2) + e^sqrt(3) = 11.765.
How do you do the second part
The second part asks for the left rule L_2 for the approximation of the integral from 0 to 2 of x^2+1. To do so, you need to divide the interval [0, 2] into two equal parts and use the left side to get the height. So the formula is L_2 = (1 - 0)*f(0) + (2 - 1)*f(1) = 1*1 + 1*2 = 3. Note that the problem statement has R_2 -- it should have said L_2.
Why is it after 10?
The function F(x) which is the integral from 0 to x of the temperature T(t) at time t can be seen in the graph as the signed area from the region between the curve and the x-axis from t = 0 to t = x. For t less than a bit before 7, the temperature is below zero and so the area you see there should be counted negatively. From that point to 10 should be counted positively as T is greater than 0. By the time one reaches 10, the area below the x-axis is still larger than the part above the x-axis. So the value of F(10) is negative. Since T is positive for times bigger than 7, F is increasing and eventually reaches zero at some time after 10.
Why is this the case?
The function is increasing and so each rectangle is entirely contained within the region between the x-axis and the graph of the curve. So, the sum of the areas of the rectangles is at most equal to the integral. Note: This assumes that the function f(x) >= 0 for all x between a and b. What can you do if the graph lies below the x-axis? What about the case where it crosses the x-axis?
Is this true because the function is decreasing instead of increasing?
That's right. The quantity R is the integral between 1 and 4 of the function whose graph is the top of the rectangles. This graph lies below the graph of 1 + 1/x because 1 + 1/x is decreasing. The difference of the integral and R is the therefore the area between the two graphs. Since areas are greater or equal to zero, the inequality is true.
How do you do this?
The curves intersect at x = 1 and x = 10 and the logarithm function lies above the line. So, you need to approximate the integral from 1 to 10 of h(x) = ln(x) - (ln(10))(x-1)/9. The rectangles have left sides at x = 1 and x = 5. So the approximation is L = (5 - 1)*h(1) + (10 - 5)*h(2). This evaluates to L = 4*(ln(1) - (ln(10))(1-1)/9 + 5*(ln(2) - (ln(10))(2-1)/9) = 2.9303.
Using x = 1, 2.5, and 5 makes the approximation L' = (2.5-1)*h(1)+(5-2.5)*h(2.5)+(7.5-5)*h(5)+(10-7.5)*h(7.5) = 2.5*h(2.5) + 2.5*h(5)+2.5*h(7.5) = 3.6763.
The antiderivative of h(x) is H(x) = x*ln(x)-x-(ln(10))*(x^2/2-x)/9 + C. So the definite integral is H(10) - H(1) = 3.6642. So the second approximation is better.
Please explain.
T_1 is the trapezoid rule with 1 trapezoid. the area of the trapezoid is f(1)/2 + f(2)/2 = 1/2 + 1/4 = 3/4 where f(x) = 1/x.
T_2 is the trapezoid rule with 2 trapezoids. The division points are at x = 1, 1.5, and 2 -- so one has T_2 = (1/2)*(f(1)/2 + f(1.5) + f(2)/2) = (1/2)*(1/2 + 1/1.5 + 1/4) = 17/24.
T_4 is the trapezoid rule with 4 trapezoids. The division points are at 1, 1.25, 1.5, 1.75, 2. The width of each rectangle is h = 1/4. So the value is T_4 = (1/4)*(f(1)/2+f(1.25)+f(1.5)+f(1.75)+f(2)/2)=.69702. These answers appear to be approaching the value of the definite integral.
Why?
From the table, one sees that the values of f(x) are decreasing whereas those of F(x) are increasing. So the function F(x) is the green one.