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Chapter 5 is about Definite and Indefinite Integrals so we worked on an exploratory problem set - section 5.1- which introduced us to these concepts.

Oil Well Problem. An oil well that is 1000 feet deep is to be extended to a depth of 4000 feet. The drilling contractor estimates that the cost in dollars per foot, f(x), for doing the drilling is f(x) = 20 + 0.000004x² where x is the number of feet below the surface at which the drill is operating. A graph of the function is shown below.

1. How much does it cost per foot to drill at 1000 ft? f(1000) = 24 so it costs $24/ft to drill at 1000 ft 4000 ft? f(4000) = 84 so it costs $84/ft to drill at 4000ft. It costs more to drill at greater depths because it is more difficult to extract the oil the further it is below the surface.

2. The actual cost of extending the well is given by the definite integral of f(x) with respect to x from x = 1000 to x = 4000. Estimate this integral by finding T(6), the trapezoidal rule sum found by dividing the interval [1000,4000] into six strips of equal width.

The Definite Integral of f on [1000, 4000] is approximately equal to

(500/2) ( f(1000) + f(4000) + 2( f(1500) + f(2000) +f(2500) +f(3000) + f(3500))

which can be evaluated as 144500, using the trapezoidal rule program.

This implies that extending the well form a depth of 1000 feet to a depth of 4000 feet would cost $144, 500.

Since the graph of function f is concave up, the trapezoidal rule approximation will overestimate the definite integral of the function f.

3. Estimate the cost again, this time estimating the integral by using rectangles whose altitudes are measured at the midpoint of each strip (that is, f(1250), f(1750), f(2250) and so on). The sum of these areas is called a RIEMANN SUM, R(6).

Cost is approximately equal to the sum of the areas of the six rectangles. Each rectangle has width 500 units and heights as described above.

Cost is approximately 500( f(1250) + f(1750) + f(2250) + f(2750) + f(3250) + f(3750)) = $143750.

This value is close to that found in question 2, using the trapezoidal rule approximation.

4. The actual value of the integral is the LIMIT of T(n), the trapezoidal sum with n increments as n becomes infinite. Find T(100) and T(500) and conjecture the exact value of the integral.

T(100) = 144,001.8 and T(500) = 144,000.072 so the exact value of the integral is probably $144,000.

5. Let g(x) be the antiderivative of f(x). (i) Find an equation of g(x). (ii) Use the equation to evaluate g(4000) - g(100) (iii) What do you notice?

(i) Thinking of f(x) as g'(x) we need to reverse the power rule for finding derivatives. The antiderivative of a constant must be that constant mulitiplied by x. The antiderivative of a quadratic term must be cubic so if g'(x) = 20 + 0.000004x² then g(x) = 20x + 0.000004x³/3 + a possible constant for which we usually use c. so, in conclusion, g(x) = 20x + 0.000004x³/3 + c (ii) g(4000) - g(1000) = 144,000 (iii) This is the same as the conjectured limit of the Trapezoidal sum as the number of increments becomes infinitely large. Maybe we have found a way to evaluate a definitel integral! HOW EXCITING IS THAT !!!!!!!!!

6. Find these antiderivatives: a. f(x) if f'(x) = 7x⁶ Answer: f(x) = x⁷ + c b. y if y ' = sinx Answer: y = -cosx + c c. u if u ' = e^(2x) Answer : u = (1/2) e^(2x) + c d. v if v '(x) = (4x+ 5)⁷ Answer : v = (4x + 5)⁸/32 + c This is a conjecture and test exercise. Think about the form of the antiderivative, then take the derivative of your conjecture and adjust coefficients accordingly. We will learn formal methods for finding antiderivatives at a later date.

Wednesday 1/06
We started with a problem of the day. In the course of working this problem we revisited, or maybe processed for the first time, the concept of a Riemann Sum. Approximating the definite integral of a function using rectangles is the basic idea behind a Riemann Sum.Specifically how the height of each rectangle is calculated determines the type of Riemann Sum; Left, Right, Mid-point or Random. In the graphic above, the rectangles with green tops represent a right Riemann sum because the height is determined by the function value of the x coordinate on the right of each sub-interval. The rectangles with the red tops represent a left Riemann sum because the height is determined by the function value of the x coordinate on the left of each sub-interval. The investigation posted on Monday used a mid-point Riemann sum, so named because the height of each rectangle is determined by the function value of the x coordinate in the center of each sub-interval.

Thursday 1/07
Today we were introduce to the differentials dx and dy which appear in the symbol for the derivative dy/dx. The slope of the tangent line to a curve at a particular point is the quotient of these differentials and we are able to use these differentials to approximate function values close to the point of tangency. We began our study of differentials using the investigation shown below. Since the slope of the tangent line is dy/dx and f '(x), we can set these two things equal to each other to find a relationship between the differentials: dy/ dx = f '(x) so dy = f '(x) dx. Also, the linearization of a function means find a linear equation which best approximates the function at a particular x value. We saw in the exploration that a tangent line approximates a function extremely well in the vicinity of the point of tangency. With this in mind, we can write a formula for the linearization of function f at the point where x = c as follows:

L(x) = f(c) + f '(c) (x - c)