Week+17+Jan+6-10

Tuesday Jan 6th
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.

Tuesday Jan 7th
Zooming in on the graph of any function will make the graph appear linear no matter how curved the graph truly is. This concept is know as local linearity. We can therefore talk about the linearization of a function which basically means can use the tangent line to a curve at a particular point to approximate the function values so long as we don't stray too far from the initial point. These ideas were explored in investigation 5.2, part of which is shown in the slideshow below. media type="custom" key="2970046" On page 3 of the investigation we discovered the relationship dy = f '(x)dx; not a new relationship but a rearranged version of the fact that dy/dx = f '(x). We are able to use this relationship to answer two types of questions: 1) If f(x) = 2cos(3x), find an expression for dy. 2) if dy = x² dx, find the antiderivative y.

The solutions to these problems are 1) If f (x) = 2cos(3x), f'(x) = -6sin(3x) so dy = -6sin(3x)dx 2) If dy = x² dx, then f '(x) = x² so we need to find the antiderivative of f(x) to find y. This implies we need to reverse the power rule - maybe add one to the power as a first attempt. So, does f(x) = x³ ? No, since f '(x) would be 3x³, three times too big. Let's try f' (x) = x³/ 3...... Perfect!

Thursday Jan 8th
The lesson today focused on the formal symbol for finding the antiderivative of a function, otherwise know as an indefinite integral. The notation for the antiderivative is a stretched out S and the expression below can be read as " the integral of f(x)dx " or the "antiderivative of f(x) with respect to x.I

In finding the antiderivative of a function, we are reversing the derivative operation so if then g '(x) = f(x).

In addition, antiderivatives are not unique. The derivative of x² + 5 is 2x and the derivative of x²+ 7 is 2x .In fact the derivative of x² + c is also 2x so long as c is a constant. Because of this, the antiderivative of any function results in a family of functions which are related to each other by a vertical shift.

Examples:

This last integral has domain issues which we will explore in depth later. At this stage, we are trying a conjecture and check method to find the antiderivative of a function. More formal techniques will be introduced in later sections.

Friday Jan 9th
Today we practiced our antiderivative skills by playing Antiderivative Bingo.(see attached file)