Antiderivatives Part III

The differential tells with respect to which variable to integrate. I am assuming the problem is:

[MATH]\int 3\sqrt{t}+1\,dt[/MATH]
Is this correct?
 
Hckyplayer8 said:
Find the indefinite integral (3 rad t + 1)dt
dt? Not like related rates, right?
Click to expand...
To Hckyplayer8, why not learn to use symbols in your posts?
If you did, it means that we can give you much faster. Why have us wait until you clarify your nonstandard abbreviations.
 
[MATH]\int (3\sqrt{t}+1)\,dt[/MATH] is solved exactly as you would solve [MATH]\int (3\sqrt{x}+1)\,dx[/MATH] except you write t instead of x
 
I think the problem lies in knowing that [math]\sqrt{t} = t^{1/2}[/math], at which point Hckyplayer8 can apply the power law that we wrote out a couple of threads ago.

-Dan
 
I seldom suggest to students (other than those concentrating in highly mathematized subjects) to learn LaTeX because they should be using their limited time on mathematics on mathematics itself rather than on an incredibly fussy rendering language. But if a student does take what I believe the sensible course and write in-line, they then must at least take the time to do that carefully. It will not be wasted; here it would have suggested the power rule.

rad t + 3 may mean t^(1/2) + 3 or (t + 3)^(1/2) as Halls pointed out. The resulting integrals are quite different.
 
Ah I'm an idiot. We literally just talked about the importance of identifying which variable we are integrating in the last thread. For whatever reason, dt just stood out to me since I just got done dealing with related rates.

HallsofIvy has the correct problem except there is an index value of 3.

Sorry for the confusion. Is there a FAQ for LaTex that I haven't seen?

As for the original problem, with the dt goof out of the way, I think I know how to solve this.

Convert the cube root of (t+1) to (t+1)1/3

Adding one to the exponent results in (t+1)4/3

Then throw the reciprocal of the exponent out in front, and lastly add the constant.

The final answer should be 3/4 (t+1)4/3 + C

Once again, thank you for the posts and continued support on here. Sorry for the confusion.
 
I am not convinced that you really know how to do problems like the one you posted. So please try the following integral and hopefully prove me wrong.
int (3t+1)(1/3)dt
 
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Oh. Am I forgetting the chain rule?

So my antiderivative was 3/4(t+1)4/3 + C

f=3/4(g)4/3
g=t+1
C = C

Which means f'=1(g)1/3 and g'=1 and C'=0

Which does not get me back to the original problem and I am wrong.
 
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Hckyplayer8 said:
Oh. Am I forgetting the chain rule?

So my antiderivative was 3/4(t+1)4/3 + C

f=3/4(g)4/3
g=t+1
C = C

Which means f'=1(g)1/3 and g'=1 and C'=0

Which does not get me back to the original problem and I am wrong.
Click to expand...
You have confirmed that Halls guessed correctly what the actual problem is, namely

[MATH]F(t) = \int \sqrt[3]{t + 1} \ dt = \int (t + 1)^{1/3} \ dt. [/MATH]
Halls suggested a u substitution. That is not necessary in this case, but it helps to prevent error.

[MATH]u = t + 1 \implies \dfrac{du}{dt} = 1 \implies du = dt.[/MATH]
[MATH]F(t) = \int \ \sqrt[3]{t + 1} \ dt = \int \ (t + 1)^{1/3} \ dt = \int \ u^{1/3} \ dt =[/MATH]
[MATH]\dfrac{1}{\dfrac{1}{3} + 1} * u^{\{(1/3) + 1\}} + C = \dfrac{\dfrac{1}{1}}{\dfrac{4}{3}} * u^{4}{3} + C = \dfrac{3u^{4/3}}{4 + C} \implies[/MATH]
[MATH]F(t) = \dfrac{3(t + 1)^{4/3}}{4}.[/MATH]
Now to check your work we use the following relationship

[MATH]F(x) = \int \ f(x) \ dx \iff f(x) = F'(x).[/MATH]
And you are correct that the chain rule comes in.

[MATH]F(t) = \dfrac{3(t + 1)^{4/3}}{4} + C = \dfrac{3u^{4/3}}{4} + C, \text { where } u = t + 1.[/MATH]
[MATH]\therefore F'(t) = \dfrac{dF}{dt} = \dfrac{dF}{du} * \dfrac{du}{dt}.[/MATH]
[MATH]F(u) =\dfrac{3u^{4/3}}{4} + C \implies \dfrac{dF}{du} = \dfrac{3 * \dfrac{4}{3} * u^{1/3}}{4} + 0 = u^{1/3}.[/MATH]
[MATH]u = t + 1 \implies \dfrac{du}{dt} = 1. [/MATH]
[MATH]\therefore F'(t) = u^{1/3} * 1 = u^{1/3} = (t + 1)^{1/3} = \sqrt[3]{t + 1}.[/MATH]
So you did the integration correctly, but messed up the check.

Whenever the chain rule comes up, u-substitution and Leibniz notation are your friends.
 
Hckyplayer8 said:
Oh. Am I forgetting the chain rule?

So my antiderivative was 3/4(t+1)4/3 + C

f=3/4(g)4/3
g=t+1
C = C

Which means f'=1(g)1/3 and g'=1 and C'=0

Which does not get me back to the original problem and I am wrong.
Click to expand...
You need to make a u-substitution. Let u=3t+1 , then du =??
Now rewrite the integral using only u's and no x's.
 
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