This week in calculus, we learned about different types of integration. The first type of substitution we learned was change of variables. This means that in the problem, there will be at least 2 different variables.
The second type we learned was area between curves. The formula for this is bSa top eq.-bottom eq.
To find a and b you have to set the equations equal to each other and solve. If the area is on the y axis, then a and b need to be y values so the equations can be solved for x. If the area is on the x axis, a and b need to be x values so the equation can be solved for y.
One of the easiest types of integration we learned this week is e integration. When dealing with e integration, whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
now you rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Another type of integration that we learned is ln integration. An example of ln integration would be:
{ 3/xdx= 3{1/xdx
3 ln |x| +C
Another type of ln integration would be:
{1/5x-1dx
u= 5x-1 du=5dx
1/5 { 5/5x-1 dx
1/5 ln |5x-1| dx +C
I finally have gotten the hang of infinite and definite integrals after taking those quizes this week. The only things i am not very good at is change in variables substitution and area between curves substitution. i understand e and ln substitution pretty well. I just do not know how to break things up in the other types of substitution and i have trouble knowing when to plug numbers in and finding where some of the numbers are coming from. If anyone can help me understand this it would be a huge help. Thanks :).
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The formula for area between curves is the integral of b (top) a (bottom) top equation - bottom equation
ReplyDeleteFirst you sketch the two graphs given to help see what exactly you are looking for.
Then you find a and b by setting the equations equal and solving
if area is \ on the y then a and b need to be y values therefore the equations must be solved for x and if a and b are x then the equations must be solved for y.
Example: Find the area of the region bounded by f(x) = 2-x^2 and g(x) = x
Set the equal to find a and b:
2-x^2 = x
x^2 + x-2=0
x= -2, x=1
The higher value will always be your b
Then depending on which equation is highest on the graph, you subtract the lower graph from the higher one.
In this case, f(x) is higher than g(x) therefore:
(2-x^2) - (x)
now integrate: 2x - 1/3 x^3 - 1/2 x^2 on 1 , -2
finally you plug in
2(1) - 1/3 (1) ^3 - 1/2 (1)^2 - [2 (-2) - 1/3 (-2)^3 - 1/2 (-2) ^2]
7/6 + 10/3 = 9/2
Wow, Chelsea is on a roll!
ReplyDeleteKudos to you, Chelsea!
Can you give me a few notes on graphing, Trina? I can only find one of my Adv Math notebooks! =/