haiku11 said:
Homework Statement
v(t) = (-2 / (1 + 2t)) + 2
I must find the antiderivative s(t) such that s(0) = 5.
The Attempt at a Solution
u = 1 + 2t du/dt = 2 dt = du/2
√(-2 / (1 + 2t dt)) + 2
If you are integrating -2/(1+ 2t)+ 2 then it is int -2/(1+ 2t)+ 2 dt. You certainly don't put the "dt" down in the denominator! Fortunately that doesn't hurt because you can hurt because you can integrate "2 dt" separately.
Oh, and why was this title "with x denominator" when there is no "x" in the problem?
=√(-2u^-1 du/2) + 2
=√(-u^-1 du) + 2
= -ln|1+2t| + 2t + C
At t = 0
-ln(1 + 2(0)) + 2(0) + C = 5
Where did this come from? Please don't add information in the middle of the problem!
-ln1 + C = 5
C = 5
s(t) = -ln|1+2t| + 2t + 5
I'm not sure if this is right, I derived it back and I got the original but there are so many things that are making me doubt myself. I have to do an antiderivative assignment without any prior knowledge so the only source I have is the internet which sometimes conflicts with itself. I read somewhere that if you have 1/x you get the special case of x^(-1 + 1) / (-1 + 1)
No, that's wrong. You do
not get that because the integral of x^n is x^{n+1}/(n+1) only if n is
not 0.
in which case it becomes ln|x|. Is this true?
"in which case"
what becomes ln|x|?
That person and that calculator are
wrong. Unfortunately, because so many text problems use only positive values for x, it is easy to forget about the absolute value. If x is not positive, ln(x) is not defined.
Problem: Find the area between y= 1/x and the x-axis for x= 1 to 2.
Answer: int_1^2 1/x dx= ln|x| evaluated between 1 and 2= ln|2|- ln|1|= ln|2|. Here, it would not have hurt to write ln x instead of ln|x| because all numbers involved are positive.
Problem: Find the area between y= 1/x and the x-axis for x= -2 to -1.
Answer: because 1/x is
below 0 for x< 0, the area is int_(-2)^(-1) (0- 1/x)dx= -int_(-2)^(-1) 1/x dx= -(ln|x|) evaluated between -2 and -1= -(ln|-1|- ln|-2|)= -(-ln(2))= ln(2). If we had not used the absolute value here, we would get no answer at all: ln(-1) and ln(-2) are not defined (as real numbers).
Also the dt is supposed to come right after each occurence of t right? Because that's what happens when you derive.
I have no idea what you are saying. What do you mean by "Because that's what happens when you derive"? Can you give an example of such differentiation? What that why you had "1+ 2t dt" in the denominator before? That is wrong. An integral is always of the form int f(x) dx no matter how complicated f(x) is.
And finally (hopefully) when do I have to use substitution like I did in this question? Is it only in situations like log (1+x) or 2(2+x) or 123/(5+2x)? And I have to encompass everything in the bracket in the substitution right?
You use a substitution when it helps and when you
can! If you set u= "some expression in x" and du is not a constant (the expression is not linear), then the derivative of u will have to
already be in the integral- you cannot just "divide" by it like you can "multiply" by the derivative using the chain rule.
For example, if the problem is to integrate (3x- 1)^5 dx, I can let u= 3x- 1, which is linear. du= 3dx so that dx= (1/3)du. The integral becomes u^5(1/3)du and I can take the constant 1/3 outside the integral.
But if the problem is to integrate (x^2- 1)^5 dx, I
cannot let u= x^2- 1. If I were to do that, du= 2xdx so dx= (1/(2x))du. But I cannot integrate u^5(1/(2x))du because I still have that "x" in the integrand. And I cannot take it outside the integral because it is a variable, not a constant.
If the problem were to integrate x(x^2- 1)^5 dx, then I
can use that substitution: u= x^2- 1 so du= 2xdx and xdx= (1/2)du. x(x^2-1)^5 dx= (x^2-1)^5(xdx)= u^5((1/2)du) and I can take that
constant, 1/2, outside the integral.
