This week in calculus we reviewed some of the concepts we did not fully understand. Let's go over some of the things we may already know:
First derivative test:
You take the derivative of the function and set it equal to zero. Then solve for the x values (critical points). Then you set the x values up into intervals between negative infinity and infinity. After that, you plug in numbers between the intervals into the first derivative to solve to see if there is a max or min or to see if the function is increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and set it equal to zero. Solve for the x values once more and set them up into intervals between negative infinity and infinity. Then plug in numbers between those intervals and plug them into the second derivative to see where the graph is concave or, concave down, or where there is a point of inflection.
Tangent lines:
The problem will give you a function and an x value. If no y value is given, plug the x value into the original function and solve for y. Once that is done, you take the derivative of the function, plug in the x value and solve to find the slope. Then set everything up into point-slope form (y-y1=slope(x-x1)).
Limit rules:
If the degree on the top is bigger than the degree on the bottom, the limit is infinity
If the degree on the top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, divide the coefficients to find the limit.
Things i'm still having trouble with:
optimization
angles of elevation and related rates
nasty looking integrals
Have a great Easter vacation :)
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The key to dealing with nasty looking integrals is to take them step by step. You will usually have to use substitution.
ReplyDeleteNumber 23 on the last AP looks really nasty because it has e's and a square root, but it is actually not that bad.
23. The integral of e^(2x) (e^x +1)^1/2 (or square root of)
Set your u equal to the inside of the square root
u=e^x+1 du=e^x
Since you have a e^(2x) and not just e^x, you have to solve u for e^x.
e^x = u-1
So now you have (u-1) (u)^1/2
The u-1 takes care of the extra e^x needed and the u^1/2 is regular substitution.
Next, distribute the u^1/2 in
u^3/2- u^1/2
Finally, integrate.
2/5 u^5/2 - 2/3 u^3/2
2/5 (e^x +1) ^5/2 - 2/3 (e^x+1) ^ 3/2 +c
Also, on the multiple choice sections, you can take the derivative of the answer choices to see which gives you the integral you are trying to find.