Jeffrey W
2/19/2016 09:25:13 pm
7.3 #15
Jeffrey W
2/19/2016 09:50:06 pm
Displacement is the numerical value given by the position function when you plug in a point in time. This means that at that moment, whatever object you are keeping track of is the output's distance away from your reference point. Displacement has a negative and positive direction
Jeffrey W
2/19/2016 09:50:51 pm
*how
Kavya Anjur
2/22/2016 09:34:59 pm
As learned in 7.1, the strategy to midweek with integrals is to first approximate what you want to find as a Reimann sum of values of a continuous function multiplied by interval lengths. With f(x) as the function and subintervals of delta x along the interval [a, b], the sums will be the approximation of a standard reimann sum. Next, a definite integral should be written of f(x) from a to b to express the limit of these sums as the norms of the partitions go to zero. The last step is to solve the integral numerically or with antiderivatives.
Kavya Anjur
2/22/2016 09:36:28 pm
**strategy to model with integrals
Kavya Anjur
2/23/2016 10:43:33 pm
For 7.2, problem #6, you can the region's symmetry to solve the problem. You would find the area of the shade region analytically by taking the area between the curves, so x^2  (2x^4), which is equal to x^2 + 2x^4. You would do two times the integral of (x^2 + 2x^4) from 0 to 1 by using the region's symmetry. If you then solve the integral by taking the antiderivative, you would get 2[(1/3)x^3 + (2/5)x^5] from 0 to 1, which is equivalent to 22/15.
Ananth Putcha
2/24/2016 06:53:16 pm
When you are finding the area of a solid with a washer as it's cross section, you take the integral of π(big radius^2)  π(small radius^2)
Ananth Putcha
3/17/2016 11:23:19 pm
Example:
Brandon C
2/26/2016 09:22:55 pm
7.2 #3 3/2/2016 05:45:54 pm
p.392 7.3 #13
Ananth Putcha
3/7/2016 10:16:32 pm
7.1 #7 v(t)=(e^sint)cost, 0≤t≤2π
Rachel Willy
3/8/2016 04:13:49 pm
Work = Force * Distance
Paige Eber
3/8/2016 09:29:17 pm
Definition; Area Between Curves
Ananth Putcha
3/8/2016 09:59:53 pm
7.2 #18 Find the area enclosed by x=y^2 and x=y+2.
Christopher Glenn
3/8/2016 11:22:32 pm
7.5 #8
Taylor Garcia
3/9/2016 09:10:27 pm
7.2 #14
Taylor Garcia
3/10/2016 06:25:49 pm
7.3 #17
Taylor Garcia
3/14/2016 09:06:31 pm
7.3 #23
Elise T
3/14/2016 11:57:12 pm
To find the volume of a solid:
Shreyas Mohan
3/15/2016 11:25:13 am
Cross Sections:
Taylor Garcia
3/15/2016 09:36:29 pm
7.3 #27
Taylor Garcia
3/16/2016 10:17:43 pm
7.3 #29
Kevin Knox
3/17/2016 11:05:54 pm
7.3 #20
Michelle H
3/18/2016 11:30:21 pm
Integrals can find:
Allison Y
3/21/2016 05:27:13 pm
7.2 #11
Allison Y
3/22/2016 07:21:22 pm
7.2 #8
Paige Eber
3/24/2016 04:51:55 pm
Occasionally we are able to use geometry to integrate, like with the function y=x. If we wanted to integrate this function over the interval [0,2] we would first recognize that the area we need to find is a right triangle with a base of 2 and a height of 2. So we can just use the formula for an area of a triangle and find the area that way
Elise T
3/25/2016 12:51:28 pm
7.2 #5
Paige Eber
3/25/2016 05:57:27 pm
7.2 #3
Shreyas Mohan
3/26/2016 08:47:03 am
Something interesting to think about is how energy and work are interrelated and how calculus can be used to prove physics topics. Most of us have learned in physics class that elastic energy is 1/2kx^2 and that F=kx. Since work is the integral of force with respect to time. W=1/2kx^2 as well, which is the same equation as energy. This shows that work is the same as a change in energy.
Kevin Knox
3/27/2016 10:02:48 pm
7.2 #1 Comments are closed.

AuthorMrs. Johnson's 20152016 BC Calculus Center for Review. By participating in this blog, you are indicating that the work that you submit is your own. If found to be otherwise true, you will not receive credit. Happy blogging!
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