January 20 Today we had a very short class, which opened with a review of how to use the calculator to evaluate integrals. For example, to evaluate the area under a curve of x^2 on the interval of 0 to 1, it would be Fnint(x^2,x,0,1)
After that, we worked on a worksheet about Accumulating Area Functions (Indefinite Integrals). Throughout this exploration, we were led to the conclusion that there is a strong relationship between the anti-derivative of a function and the integral of the function.
Today we learned about the Fundamental Theorum of Calculus. The first part of the FTC states that the rate of change of area, is the same as the height of the function f at that instant. The Second part of the FTC is a shortcut to find the area under the curve, for definite integrals. (A definite integral is a number, an indefinite integral is a function.) Before Newton, this was very hard and took a very long time to find. Now, though, we can just find the anti-derivative of the function, plug in the top number, plug in the bottom number, and subtract.
The chief discovery of today's class was that the graphs of Accumulating Area functions can be evaluated by using the graph of the function being integrated. Specifically, the values of minimum and maximum points and points of inflection of an A.A. function can be determined from the graph of the base function. Where the graph of f(x) goes from - to +, there is a minimum on the A.A. graph, likewise, when f(x) goes from + to -, there is a max on the graph of the A.A. function. Points of inflection occur when the slope of f(x) changes signs (- to + or vice versa). Instantaneous slopes of A.A. functions can also be found by looking at the graph of f(x)'s y-value at that point.
January 23 Today we received a worksheet that showed us how to distinguish between the meaning of certain integrals. For example, the integral of the rate of change of a population for 1 year is the net change of the population for that year. When we add the initial population to that, we get the total population after 1 year.
Then we got another worksheet that explained the difference between the value of the integral and the total area. On the first problem, we found that the value of the integral was 0 because half of the graph was above the x-axis (1.33) and half was below (-1.33), which canceled out to get 0. The total area underneath the graph, however, was the total of the first half and the second half (1.33+1.33=2.66) which we can represent by adding the absolute value sign to the function inside the integral. That way both the halves are flipped to be above the x-axis.
The second problem showed that net change of the bug's position was the distance to the right minus the distance to the left. If we wanted to get the total distance traveled, however, we would need to do the same thing as before and add the absolute value sign to the function inside the integral.
First we talked about: Normally when you have v(t) representing the instantaneous velocity of an object, the integral from time one to time two of v(t) dt represents displacement, which is kind of like where you end up. However, the integral of the absolute value of v(t) represents total distance because everything that was negative is counted as positive. A good general examply of distance vs. displacement is:
If you traveled 8 units to the right, and two to the left, the total distance would be 10 and the displacement would be 6.
Next we talked about how before theinvention of the Fundamental Theorem of Calculus, the integral symbol stoof for the limit as n approaches infinity of a Reimann Sum. After this revolutionary discovery, the integral symbol became associated with finding an anti derivative
When we have a definite integral we know that the answer is going to be a number, because we have certain endpoints and we can calculate a certain numerical value. When we have an indefinite integral, our answer is in the form of a function because we are antideriving a function.
We also learned some of the rules for indefinite integrals. You must rewrite the integral into some form of addition or subtraction before you anti-differentiare. If we have the integral of (1-t)(2+t^2) we can not anti-derive each separately. We must distribute to get the integral of (2 + t^2-2t-t^3) and then anti-derive. Remember that we cannot forget to add +C because there could be a constant.
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January 20
Today we had a very short class, which opened with a review of how to use the calculator to evaluate integrals. For example, to evaluate the area under a curve of x^2 on the interval of 0 to 1, it would be
Fnint(x^2,x,0,1)
After that, we worked on a worksheet about Accumulating Area Functions (Indefinite Integrals). Throughout this exploration, we were led to the conclusion that there is a strong relationship between the anti-derivative of a function and the integral of the function.
Today we learned about the Fundamental Theorum of Calculus. The first part of the FTC states that the rate of change of area, is the same as the height of the function f at that instant.
The Second part of the FTC is a shortcut to find the area under the curve, for definite integrals. (A definite integral is a number, an indefinite integral is a function.) Before Newton, this was very hard and took a very long time to find. Now, though, we can just find the anti-derivative of the function, plug in the top number, plug in the bottom number, and subtract.
The chief discovery of today's class was that the graphs of Accumulating Area functions can be evaluated by using the graph of the function being integrated.
Specifically, the values of minimum and maximum points and points of inflection of an A.A. function can be determined from the graph of the base function. Where the graph of f(x) goes from - to +, there is a minimum on the A.A. graph, likewise, when f(x) goes from + to -, there is a max on the graph of the A.A. function. Points of inflection occur when the slope of f(x) changes signs (- to + or vice versa).
Instantaneous slopes of A.A. functions can also be found by looking at the graph of f(x)'s y-value at that point.
January 23
Today we received a worksheet that showed us how to distinguish between the meaning of certain integrals. For example, the integral of the rate of change of a population for 1 year is the net change of the population for that year. When we add the initial population to that, we get the total population after 1 year.
Then we got another worksheet that explained the difference between the value of the integral and the total area. On the first problem, we found that the value of the integral was 0 because half of the graph was above the x-axis (1.33) and half was below (-1.33), which canceled out to get 0. The total area underneath the graph, however, was the total of the first half and the second half (1.33+1.33=2.66) which we can represent by adding the absolute value sign to the function inside the integral. That way both the halves are flipped to be above the x-axis.
The second problem showed that net change of the bug's position was the distance to the right minus the distance to the left. If we wanted to get the total distance traveled, however, we would need to do the same thing as before and add the absolute value sign to the function inside the integral.
First we talked about:
Normally when you have v(t) representing the instantaneous velocity of an object, the integral from time one to time two of v(t) dt represents displacement, which is kind of like where you end up. However, the integral of the absolute value of v(t) represents total distance because everything that was negative is counted as positive. A good general examply of distance vs. displacement is:
If you traveled 8 units to the right, and two to the left, the total distance would be 10 and the displacement would be 6.
Next we talked about how before theinvention of the Fundamental Theorem of Calculus, the integral symbol stoof for the limit as n approaches infinity of a Reimann Sum. After this revolutionary discovery, the integral symbol became associated with finding an anti derivative
When we have a definite integral we know that the answer is going to be a number, because we have certain endpoints and we can calculate a certain numerical value. When we have an indefinite integral, our answer is in the form of a function because we are antideriving a function.
We also learned some of the rules for indefinite integrals. You must rewrite the integral into some form of addition or subtraction before you anti-differentiare. If we have the integral of (1-t)(2+t^2) we can not anti-derive each separately. We must distribute to get the integral of (2 + t^2-2t-t^3) and then anti-derive. Remember that we cannot forget to add +C because there could be a constant.
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