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Monday 1/11
Today we worked some problems on linearization of a function and calculated values of the differential dy. These problems are shown below.

We then were given the following practice problems. Find an equation for the differential dy in each case. 1) y = (x³ + 2)⁸ Ans: dy = 24x²(x³ + 2)⁷dx 2) y = x²tanx Ans: dy = (x²sec²x + 2xtanx)dx 3) y = ln(secx) Ans: dy = (1/cosx)secxtanxdx =sec²xtanx =dx

Find an equation for the indefinite integral :

1) dy = 8x³ dx Comparing this to dy = f '(x)dx leads us to think f '(x) = 8x³. Now we need to think of function f, the antiderivative of 8x³. The function whose derivative is x³ is some coefficient multiplied by x⁴. First conjecture is that f(x) = x⁴ but if this were the case, f '(x) = 4x³ not 8x³. Second conjecture is f(x) = 2x⁴ ; certainly the derivative of this function is 8x³ but so would the derivative of 2x⁴ + 1, or 2x⁴ + 3 or 2x⁴ +72.... This implies that the antiderivative of a function is not unique; it is a family of functions connected to each other by a vertical shift. Final answer: f(x) = 2x⁴ + c where c is referred to as the constant of integration.

2) dy = (3x+2)³dx Ans: y = (3x+2)⁴/12 + c 3) dy = e^(sinx)cosxdx Ans: y = e^(sinx) +c 4) dy = tanx sec²xdx Ans: y = (tan²x) /2 + c

Tuesday 1/12
We first went over the solution to problem #5 from the homework, the solution is posted below. We didn't go over number 6 but the solution is shown here.

Wednesday 1/ 13
Today we used the integral symbol for the first time. The symbol is an elongated S which could make sense if we think of the relationship suggested in Exploration 5.1; that is, there is a connection between the definite integral of a function, calculated as an area under a curve using the sum of rectangles or trapezoids, and the antiderivative of that same function. Today we literally practiced some antiderivatives- problems shown in the graphics below.



The derivative of tanx is sec²x so the antiderivative of sec²x must be tanx. The derivative of tan(3x) is 3sec² (3x) so the antiderivative of sec² (3x) must be tan(3x) /3 + c

Rewrite the integrand as 4e^(-2t). The derivative of e^(-2t) is -2e^(-2t) so the antiderivative of -2e^(-2t) is e^(-2t). We have the antiderivative of 4e^(-2t) which is -2 times of -2e^(-2t). Conclusion: Antiderivative of 4e^(-2t) must be -2e^(-2t) +c

HAPPY HOMEWORK! Page 202 # 1-31 odd. Lots of anti-derivatives!

Thursday 1/14
Riemann Sums and evaluating a definite integral is how we spent our day today! We know that a definite integral of a function is the area between the function and the x axis and the two vertical lines represented by the upper and lower bounds of the integral. These ideas are explored and explained in the graphics which follow. There are three main types of Riemann Sum: 1) Left Riemann Sum Heights of the rectangles are determined by the function values on the left of each sub-interval.

2) Right Riemann Sum Heights of the rectangles are determined by the function values on the right of each sub-interval.

3) Mid-point Riemann Sum Heights of the rectangles are determined by the function values at the midpoint of each sub-interval.

See the next graphic for an example of a right Riemann Sum.



a) Definite Integral is approximately equal to: 1 ( f(1) + f(2) + f(3) + f(4) ) = 2 + 5 + 10 + 17 = 34

b) Using the TRAP RULE program we found the following results: __Number of Trapezoids__ __Definite Integral__ 4 34 10 25.44 50 25.3376 100 25.3344 We could conjecture that the value of the definite integral may be 25.33333333

Notice that g(4) - g(0) is the same value as our conjectured value of the definite integral!

From here we thought about trying to find the exact value of a definite integral by placing n rectangles within the defined region. If we can find an expression for the definite integral in terms of n, allowing n to approach infinity should, conceptually, result in the exact value we are looking for. This idea is presented in the next graphic- a linear function was used to simplify the algebra.

We each chose a function and attempted to find the definite integral of our function over the interval [0,2], making sure that the function we chose had positive y values over the interval [0,2]. Later on we will deal with issues such as "what happens if the function dips below the x axis?". Here are some more examples: media type="custom" key="5197637"

Friday 1/15
Today we continued with the idea of finding the limit of a Riemann Sum, this time using an interval whose left bound was not zero! Unfortunately I do not have any graphics to share- check back later!