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U substitution on ∫sin(x^5)dx

  • #1

Homework Statement



Use Part 2 of the Fundamental Theorem of Calculus to find the derivative.

[tex] \int_3^x sin(t^{5}) \, dt [/tex]


Homework Equations





The Attempt at a Solution



I know the general idea of what I'm supposed to do as far as evaluate the indefinate integral and then do a subtraction of the upper limit and lower limits...but I can't even get to the point of finding the indefinate integral. (maybe it's that I just "think" I know what i'm supposed to do..)

[tex] \int_3^x sin(t^{5}) \, dt [/tex]

I'm letting u = [itex]t^{5}[/itex]

so du = [itex]5t^{4}[/itex]

then it looks like dt can be replaced by [itex]\frac{1}{5t^{4}}[/itex]

so that [tex] \int_3^x \frac{1}{5t^4} sin(u) \, du [/tex]

However, our professor has instructed that mixing variables within the integral is not allowed because it can't be evaluated.

So how to I do u-substitution on this integral?
 

Answers and Replies

  • #2
vela
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You don't. Note that the problem says to use the fundamental theorem of calculus. Don't ignore that bit of info.
 
  • #3
SammyS
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Homework Statement



Use Part 2 of the Fundamental Theorem of Calculus to find the derivative.
[tex] \int_3^x sin(t^{5}) \, dt [/tex]

Homework Equations



The Attempt at a Solution



I know the general idea of what I'm supposed to do as far as evaluate the indefinate integral and then do a subtraction of the upper limit and lower limits...but I can't even get to the point of finding the indefinate integral. (maybe it's that I just "think" I know what i'm supposed to do..)
[tex] \int_3^x sin(t^{5}) \, dt [/tex]
I'm letting u = [itex]t^{5}[/itex]

so du = [itex]5t^{4}[/itex]

then it looks like dt can be replaced by [itex]\frac{1}{5t^{4}}[/itex]

so that [tex] \int_3^x \frac{1}{5t^4} sin(u) \, du [/tex]

However, our professor has instructed that mixing variables within the integral is not allowed because it can't be evaluated.

So how to I do u-substitution on this integral?
What vela said, plus ... you're to find the derivative.
 
  • #4
oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day. And so I figured the fundamental theorem being referred to was the property that is actually FTC1. Anyway, Thanks :) sin(x^5) duh. I turned a 30 second problem into a 20 minute headache. Thanks Vela and Sammy.
 
  • #5
SammyS
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oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day. And so I figured the fundamental theorem being referred to was the property that is actually FTC1. Anyway, Thanks :) sin(x^5) duh. I turned a 30 second problem into a 20 minute headache. Thanks Vela and Sammy.
The limits od the integral are 3 and x, so the integral is a function of x, not t.

The derivative of this integral is with respect to x.

I.e.

Find
[itex]\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)\ .[/itex]​
 
  • #6
The limits od the integral are 3 and x, so the integral is a function of x, not t.

The derivative of this integral is with respect to x.

I.e.

Find
[itex]\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)\ .[/itex]​

Thanks, I noticed that right after I submitted the post and editted with the correct notation right afterwards.
 
  • #7
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Is this just [itex] 5x^4 sin(x^{5}) [/itex] ?
 
  • #8
SammyS
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Is this just [itex] 5x^4 sin(x^{5}) [/itex] ?
No.

The anti-derivative of [itex] 5x^4 \sin(x^{5}) [/itex] is [itex] -\cos(x^{5}) +C[/itex],

not [itex]\displaystyle \int \sin(x^{5}) dx\ .[/itex]
 
  • #9
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So this should be [itex] sin(x^{5}) [/itex], correct?
 
  • #10
HallsofIvy
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No, it isn't. What is the derivative of sin(u(x)) with respect to x. What is the derivative of cos(u(x)) with respect to x?
 
  • #11
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cos(u(x)) u'(x) and -sin(u(x)) u'(x).
 
  • #12
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So this should be [itex] sin(x^{5}) [/itex], correct?
This result looks correct to me. There is no x under the integral sign. I don't understand what Halls of Ivy is saying.

Chet
 
  • #13
SammyS
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So this should be [itex] sin(x^{5}) [/itex], correct?
If you mean:
Is [itex]\displaystyle
\frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\sin(x^5)\ ?[/itex]​

Then I agree with Chestermiller that, "Yes it is."

Here's how I arrive at that:

Let F(t) be an anti-derivative of sin(t5). In other words, let [itex]\displaystyle F(t)=\int sin(t^{5}) \, dt\ .[/itex]

Then [itex]\displaystyle \int_3^x sin(t^{5}) \, dt=F(x)-F(3)\ .[/itex]

Therefore, [itex]\displaystyle
\frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\frac{d}{dx}\left(F(x)-F(3)\right)=\sin(x^5)-0\ .[/itex]
 
  • #14
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oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day.
"When all else fails, read (or reread) the instructions."
 

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