- Using U-substitution to evaluate integrals is difficult to set up. It is sometimes hard to know which value will be the U and which value will be the DU.
- Finding the area between two curves is challenging when I forget to find which function is on top and when there are more than two places where the functions intersect.
- Finding the volume of a region using cylindrical shells is difficult because I often confuse this method with the washer method.
- Finding the volume of a region using the washer method is difficult because I often confuse this method with cylindrical shells.
- Using the limit definition to find the area under the curve also proves to be difficult because I do not remember the limit definition.
2. Do the Problems
- U-SUB
- Find the definite integral of x(1 - 3x2)4dx u=1-3x^2, du=-6xdx, -1/30(1-3x^2)^5 +c
- Find the definite integral of (1+sec(x))2(sec(x)tan(x))dx u= 1 + sec, du= sec(piX)tan(piX)dx, ((2-3x)^3)/3 +c
- Find f(x) when f(x) = x((2-3x)^1/2)dx u= 2-3x, du=-3dx, (2/45(2-3x)^5/2)(-4/27(2-3x)^3/2)+c
- Cylindrical Shells
- Find the volume of the region bounded by y = 2x3, y=2x, x = 1, and x = 2;
- revolved about the y- axis 2pi from 0 to 1 x(2x^3-2x)dx, 232/15pi
- Washer Method
- Find volume of area bounded by x = 0, y = x +4 and y = 12x2 revolved around x = -1 pi from -1 to 0, (x+4)^2 -(1/2x^2)^2dx, 737/60pi
- Limit Definition
- Evaluate the following using the limit definition and using integrationf(x) = 4x - 5 on [1, 5] The Summation of 4/n 4(1+4i/n) -5, The answer is 28.
- Area Between the Curves
- Find area of the region bounded by : f(x) = 2x-2 g(x) = -x2-2x+3 Interval -5 to 1, (-x^2-2x+3) - (2x-2), 108/3
- Find area of the region bounded by : f(x) = 2x3-1, g(x) = 2x-1 Interval 0 to 1, (2x-1)-(2x^3-1) The answer for this half is one half. We must multiply by to because the function is cubic and symmetrical across the y-axis. The answer is just one.
3. Create a problem that is similar.
- U-Sub
- Y= sec^2(x)tan^3(x), U= tan(x), Du= sec^2(x)dx, 1/4tan^4(x)+c
- Cylindrical Shells
- y=x-x^2, y=0, about x=2, 2pi integral of 0 to 1, (2-x)(x-x^2)dx, pi/2
- Washer Method
- y=x, y=x^2 about the x-axis. pi integral of 0 to 1, ((y^1/2)^2)-(y)^2dy, pi/6
- Limit Definition
- 2x^2-1 on [1, 5] limit as n approaches infinity of the summation of 4/n 2(1 + 4i/n)^2 -1, 236/3
- Area Between the Curves
- y= x^2-2x-4, y=x Integral -1 to 4, (x) - (x^2-2x-4)dx, 125/6
I like how you organized your work in correspondence with the directions of the blog post. It makes your work easy to follow. However, I wish the title of your blog post was a bit less mainstream.
ReplyDeleteWhen you find the area between the curves, I think you're supposed to do the calculus for both pieces instead of just multiplying by two, even though you got the same answer
ReplyDeleteThat is correct, Maddie. In this case though, you can multiply by two because of the symmetry.
ReplyDeleteI like how you did not share this with me.... and forced me to search for R McCune's Calculus Blog.
ReplyDelete