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Math Help - derivative of an integral

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    derivative of an integral

    I'm struggling with this: \frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt

    Using \frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x), I get \frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}, which is wrong (I think).
    Last edited by th%$&873; January 14th 2008 at 03:26 PM. Reason: missing derivative sign
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    Quote Originally Posted by th%$&873 View Post
    I'm struggling with this: \frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt

    Using \frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x), I get \frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}, which is wrong (I think).
    yes that is wrong.

    apply a minus sign to change the limits: \frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt = - \frac{d}{dx}\int_{\pi / 2}^{x} sin\frac{t}{2}cos\frac{t}{3} \, dt

    now apply the second fundamental theorem of calculus (the rule you stated)
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    Quote Originally Posted by th%$&873 View Post
    I'm struggling with this: \frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt

    Using \frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x), I get \frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}, which is wrong (I think).
    Get ready to kick yourself, sport.

    \frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt = - \frac{d}{dx}\int_{\pi/2}^{x} sin\frac{t}{2}cos\frac{t}{3} \, dt .

    And you know that \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \, , right?

    doh! Beaten by Jhevon. But my answer's funnier
    Last edited by mr fantastic; January 14th 2008 at 03:32 PM. Reason: Added the last line
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    ...
    doh! Beaten by Jhevon.
    we have a "doh!" emoticon you know:

    (the emoticons were a lot better when they were animated)

    But my answer's funnier
    working on getting nominated for the "Funniest Member" award for 2008 i see
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    we have a "doh!" emoticon you know: Mr F says: yes, I know. But the emoticons aren't easily accessible (as far as I can see) when you edit a post.

    (the emoticons were a lot better when they were animated)

    working on getting nominated for the "Funniest Member" award for 2008 i see
    To paraphrase the old sporting cliche, I'm just taking it one post at a time

    (And if there was a "Grouchiest Member" award, somtimes I think I could easily lose my New Year resolution resolve )
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    apply a minus sign to change the limits
    I thought that's what I did--which is why I got \frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2} instead of -\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}+\frac{cos\f  rac{x}{3}cos\frac{x}{2}}{2}
    And you know that \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \,, right?
    Yes. But why does my calculator give me -\frac{sin\frac{5x}{6}}{2}-\frac{sin\frac{x}{6}}{2}?
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    Quote Originally Posted by th%$&873 View Post
    I thought that's what I did--which is why I got \frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2} instead of -\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}+\frac{cos\f  rac{x}{3}cos\frac{x}{2}}{2}


    Mr F says: No no no! You do NOT differentiate f(t) \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \, NOT f'(x)!!


    Yes. But why does my calculator give me -\frac{sin\frac{5x}{6}}{2}-\frac{sin\frac{x}{6}}{2}? Mr F says: Without seeing what your input is, I really couldn't say. But there's an old computer programming adage that has never been truer when it comes to CAS calculators: Garbage in, garbage out. No offence intended
    ..
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    ...
    (And if there was a "Grouchiest Member" award, somtimes I think I could easily lose my New Year resolution resolve )
    sorry to burst your bubble, but there are members here who are a lot more grouchy than you are. you may get nominated, but you wouldn't get a single vote...well, maybe if rcmango voted
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    sorry to burst your bubble, but there are members here who are a lot more grouchy than you are. you may get nominated, but you wouldn't get a single vote...well, maybe if rcmango voted
    Aha yes, but you know I'm holding back My resolve for my New Year resolution stands resolute (currently anyway )
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    Quote Originally Posted by mr fantastic View Post
    Mr F says: No no no! You do NOT differentiate f(t)   \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \, NOT f'(x)!!
    But Jhevon said
    now apply the second fundamental theorem of calculus (the rule you stated)
    I would have simply substituted x like you said, mr fantastic, but my calculator said otherwise, so I changed course.
    Without seeing what your input is, I really couldn't say. But there's an old computer programming adage that has never been truer when it comes to CAS calculators: Garbage in, garbage out. No offence intended
    Believe me, I am acutely aware of what happens when the input into a calculator isn't formatted right. I entered it exactly as shown in my posting, checked and re-checked. Of course I'm not expecting you to troubleshoot my calculator.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by th%$&873 View Post
    But Jhevon said
    I was referring to:
    Quote Originally Posted by th%$&873 View Post
    Using \frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)
    that is the right rule to use. it is a variation of the second fundamental theorem of calculus (which mr fantastic gave you the formula for) which is derived by using the chain rule. i was not in anyway referring to the format of your answer. in fact, i was going to ask you in my first post why is it you have two terms as opposed to one
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    got it.
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  13. #13
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    Quote Originally Posted by th%$&873 View Post
    But Jhevon said
    I would have simply substituted x like you said, mr fantastic, but my calculator said otherwise, so I changed course.

    Believe me, I am acutely aware of what happens when the input into a calculator isn't formatted right. I entered it exactly as shown in my posting, checked and re-checked. Of course I'm not expecting you to troubleshoot my calculator.
    If you have a TI-89, the input is:

    d( \int(sin(t/2)*cos(t/3),t,x, \pi/2),x)

    and you get ....... ohoho! ..... and you get -\frac{\sin\frac{5x}{6}}{2}-\frac{\sin\frac{x}{6}}{2} ..... now that's interesting .....

    But remember that \sin(A + B) + \sin(A - B) = 2 \sin A \cos B.

    So here's a question for you: What happens if you let A = -\frac{x}{2} and B = \frac{x}{3} in the above identity?

    Hint: Stamp this case as solved.

    (Or perhaps not ...... why A = -\frac{x}{2} rather than A = \frac{x}{2}...... ? Why?)

    What's the moral of the story? (The answer has the words "unexpected" and "output" in it). Hmmmm .... the CAS calculator works in mysterious ways sometimes.
    Last edited by mr fantastic; January 14th 2008 at 07:01 PM. Reason: Added the "perhaps not" bit
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    Quote Originally Posted by th%$&873 View Post
    \frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)
    And just to seal the deal and end this thing, what does this rule boil down to when g(x) = x?
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    But remember that \sin(A + B) + \sin(A - B) = 2 \sin A \cos B.

    So here's a question for you: What happens if you let A = -\frac{x}{2} and B = \frac{x}{3} in the above identity?

    Hint: Stamp this case as solved.

    (Or perhaps not ...... why A = -\frac{x}{2} rather than A = \frac{x}{2}...... ? Why?)
    My trig is a little rusty.

    What's the moral of the story? (The answer has the words "unexpected" and "output" in it). Hmmmm .... the CAS calculator works in mysterious ways sometimes.
    That's why I asked you what was going on, because sometimes calculators give odd, but correct results. If you're good at math, you can see what happened. Sometimes the boolean evaluator function doesn't work, and I have to compare two expressions numerically (maybe there's a better way). Bottom line is, I thought this problem was more complicated than it actually was, and got confused.

    Quote Originally Posted by mr fantastic View Post
    And just to seal the deal and end this thing, what does this rule boil down to when g(x) = x?
     g'(x) <br />
is 1, so it would be f(x)?
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