Post comments here relevant to Chapter 5 only.
All continuous functions are integrable. That is, if a function f is continuous on an interval [a,b], then it's definite integral over [a,b] exists.
Basically the important thing to realize is if a function is continuous you can take the integral. It doesn't have to be differentiable to integrate it.
Gives u the average value of a function, f(x), if it's integrable on [a,b].
Similar to my last post about average value but incorporating MVT into it.
Don't forget u^2 because we are calculating area!
The Trapezoidal Rule is used to more accurately estimate the area under a curve. It is equal to (LRAM + RRAM)/2. Keep in mind that it is only an estimation of the area. It won't get you the exact area, but it will get you very close.
By graphing it, we can tell this function makes a triangle on the given interval. The area of a triangle is equal to (1/2)(Base)(Height).
Looking at the graph we can tell the triangle has a base of two (abs(-1) +1 = 2) and a height of 1 since the triangle is isosceles and 1 is the length of the perpendicular bisector of the triangle.
Therefore, the area of the function from -1 to 1 is equal to the area of the triangle formed by the function, or (1/2)(2)(1), which equals 1.
By transforming the equation we find that it is a circle with a radius of 4 and centered around the origin. However, due to the initial equation, we are restricted to only the top half of the, circle located in quadrants 1 and 2.
Since the circle is centered at the origin and has a radius of 4, plus only the top half of the circle fits the original equation, with the bounds we are given means we are taking the area of a quarter of the circle.
The area of a circle is πr^2, and we are taking a quarter of that, which gives us a new equation: (1/4)πr^2.
Graphing this function, we find that it makes a corner centered at the origin. If we create an imaginary VASY at the origin, we can see two distinct triangles, both with heights above the x-axis. That means to find the area of the graph we need to find the area of the two triangles and add them together.
Looking at our function and bounds, we can see the triangle on the left has a base of 2 and a height of 2 as well. The triangle on the right has a base and height of 1.
Sketching the graph, we find that we are given a linear line that has a positive slope. If we set up our bounds and shade within them on our graph we find that we have a trapezoid looking shape.
The area of a trapezoid is equal to ((A+B)/2)*h where A and B are the varying base lengths and h is the height of the trapezoidal figure perpendicular to the parallel bases.
Looking at our graph we find that bases created by the x-axis and our graphed function aren't parallel and won't work for our area equation. To solve this we flip the figure on its side and use the normally vertical lines extending straight down from our bound points as horizontal bases.
In the original graph we found our bound points to be (-2,2) and (4,5), this means the length of these two bases are 2 and 5 respectively. We can also use the original x-axis as our height measurement, for a height of 6.
Using the power rule, we find the anti-derivative to be 7x. 7 is a constant, so it technically has an x component of x^0. The power rule states that we add one to the power of x and divide the coefficient by the new power. So the 0+1 = 1, and 7/1=7. This creates the new term of 7x.
Going back to page 159 of our book, we see that the derivative of tan^-1(u) is (1/(1+u^2)). Since Anti-derivatives start with the derivative and end with the normal function, we can tell the Anti-derivative of our original problem is tan^-1(x).
Using the fundamental theorem of calculus part two, we evaluate our anti-derivative using our bounds.
Find the average value of the function y=x^3 + 1 on the interval [0,2].
The average value of the function on the interval [0,2] is 3.
First step is to find the anti-derivative of our integral. Using the power rule, where each term has its variable power raised by one and divided by that new power, we find that 2t becomes t^2, because t^(1+1) = 2, and 2/2 =1, and 3 becomes 3t because t^(0+1) = 1, and 3/1=3. So our anti-derivative is t^2-3t.
Using the fundamental theorem of calculus part two, we substitute our bounds into our anti-derivative. The theorem also states that to find the area we must subtract the upper bound from the lower bound.
The anti derivative of 5x is (5/2)x^2 using the power rule.
Using FTC pt 2, we evaluate the integral using the upper bound subtracted by the lower bound.
Mrs. Johnson's 2015-2016 BC Calculus Center for Review.
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