Section 4.10: Antiderivatives
Wassup!
Section 4.10 is all about antiderivatives(Precisely how to find antiderivatives).
Before any specifics, the class should know what the antiderivative exactly is.
-Antiderivative(Book's interpretation): F(x) is an antiderivative of f(x) if F'(x)=f(x), on an interval
-Antiderivative(My interpretation): Opposite of the derivative(The class finds the original function, given the derivative of that function)
Given the definition of the antiderivative, a theorem representing the general equation for antiderivatives can be developed: Antiderivative of the given function=F(x)+C, C being the constant
Example:f(x)=3x^4
F(x)=(3x^5)/5+C, which means F(x)=3(x^5)/5+10000,F(x)=3(x^5)/5+573892
Finding the antiderivative(Finding the original function, given the derivative) is just as easy as finding the derivative.
Simply remember the rule: x^n=(x^n+1)/(n+1), when n does not equal -1
REMEMBER that this rule can be applied to any term in a function
Example:f(x)=5x^4+2x^3-3x^2
F(x)=(5x^5)/5+(2x^4)/4-(3x^3)/3=x^5+(2x^4)/4-x^3 + C
Example:f'(x)=5x^4+2x^3-3x^2, f(0)=2(MADE UP, MAY NOT WORK OUT EVENLY)
f'(x)=x^5+(2x^4)/4-x^3+C, C=1+2-1+2=4(WHY? Add the coefficients of the other terms plus what is given[In this case, coefficients are 1,2,-1, and given of 2])
However, the rule: x^n=(x^n+1)/(n+1) has a problem when n=-1, since the antiderivative ends up undefined(Since the denominator=0).
Example: f(x)=x^-1
F(x)=(x^0)/0=Undefined
Luckily, there is another rule that allows us to find the antiderivative of a function when n=-1. The rule: ln x, when n=-1
Example: f(x)=x^-1=1/x
F(x)=ln x + C
Here is a link that can help you understand antiderivatives even more!
http://v5o5jotqkgfu3btr91t7w5fhzedjaoaz8igl.unbsj.ca/~talderso/UWOnotes/UNIT2.pdf
Now, before I leave...
Here are some pictures of some my favorite breakdance crews/breakdancers
BBoy Hong 10
BBoy Wake-Up
Expression Crew
REMEMBER, YOUR NEXT KANE!!!
2 Comments:
i may be misunderstanding your explanation for your example, but i think the constant, c, is 2. if you plug in 0 for the new function, you get 2 for the constant, not 4. I do not understand why you would add the coefficients together. please correct me if im wrong, which i very well may be (im trying to study for the test)
Joey's correct - the constant should be 2, not 4. Joey's method of calculating the constant is the correct method...
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