Integration and antiderivatives in class

[3ephemeralwnd]

Ive just started learning about integration and antiderivatives in class, and I've got a question

Say we have:

f(x) = 1/x^3 , and f'(x) = F(x)

g(x) = 1/X , and g'(x) = G(x)

then to find the antiderivative of f(x), i would solve it like this:
first rewrite it as : x^(-3),
= x^(-3 + 1) / ( -3+1)
= x^ (-2) / (-2)
which i believe is the correct antiderivative function

but for g(x), if i were to use the same method, i'd get
x^(-1),
= x^(-1+1)
= x^0
= 1
but the actual antiderivative is ln|x| (since the derivative of lnx = 1/x)

why doesn't that the first method work anymore?

 

[lugita15]

Ive just started learning about integration and antiderivatives in class, and I've got a question

Say we have:

f(x) = 1/x^3 , and f'(x) = F(x)

g(x) = 1/X , and g'(x) = G(x)

then to find the antiderivative of f(x), i would solve it like this:
first rewrite it as : x^(-3),
= x^(-3 + 1) / ( -3+1)
= x^ (-2) / (-2)
which i believe is the correct antiderivative function

but for g(x), if i were to use the same method, i'd get
x^(-1),
= x^(-1+1)
= x^0
= 1
but the actual antiderivative is ln|x| (since the derivative of lnx = 1/x)

why doesn't that the first method work anymore?

Actually, you'll get x^(-1+1)/(-1+1) which is undefined because the denominator is zero. All this shows is that the power rule is inapplicable in this case and another method has to be applied. Depending on how you define the natural logarithm, it's either trivial to find the antiderivative of x^-1 or it will take some work, for instance if you start with the infinite series definition of e.

 

[Mark44]

Ive just started learning about integration and antiderivatives in class, and I've got a question

Say we have:

f(x) = 1/x^3 , and f'(x) = F(x)

The above is incorrect if F(x) is supposed to be an antiderivative of f(x). As you have things, F(x) = f'(x) = d/dx(x^-3) = -3x^-4. Since the thread title indicates that you are interested in antiderivatives, not derivatives, the second equation above should be f(x) = F'(x). IOW, the derivative of some as-yet-unknown function F(x) is 1/x³.

g(x) = 1/X , and g'(x) = G(x)

Same as above. What I think you mean is that g(x) = G'(x), and you would like to find G(x).

then to find the antiderivative of f(x), i would solve it like this:
first rewrite it as : x^(-3),
= x^(-3 + 1) / ( -3+1)
= x^ (-2) / (-2)

For this problem, you have the basic technique right, but what you are saying isn't correct. The main thing that is wrong is that you are saying that the f(x) is equal to its own antiderivative, which is almost never true. Here's what you should be saying:

f(x) = x^-3
F(x) = \int f(x) dx = \int x^{-3}dx = \frac{x^{-2}}{-2} + C

The other thing is that you are leaving out the constant of integration, which I show above as C. Differentiation and antidifferentiation are opposite operations, with one important difference: For any given (differentiable) function, there is only one derivative. For any given (integrable) function, there are an infinite number of antiderivatives.

which i believe is the correct antiderivative function

but for g(x), if i were to use the same method, i'd get
x^(-1),
= x^(-1+1)
= x^0
= 1
but the actual antiderivative is ln|x| (since the derivative of lnx = 1/x)

why doesn't that the first method work anymore?

You are using the integration power rule, but you aren't reading the fine print. This rule says that
\int x^n dx = \frac{x^{n +1}}{n + 1} + C
provided that n \neq -1.
That's the fine print.

 

[HallsofIvy]

Specifically, x^{n+1}/(n+1) is the anti-derivative of x^n (as long as x\ne -1 because the derivative of x^n is nx^{n-1} for all n and so the derivative of x^{n+1}/(n+1)<formula class="mathJaxEqu" data-display="false">is <formula class="mathJaxEqu" data-display="false">(n+1)/(n+1)x^{n+1-1}= x^n[/math].<br /> <br /> That is NOT correct for n= -1 because the derivative of <formula class="mathJaxEqu" data-display="false">x^{-1+ 1}= x^0= 1</formula> is 0.<br /> <br /> However, there <b>is</b> a function, f(x), whose derivative is 1/x. Do you know what it is?</formula></formula>