Results 1 to 7 of 7

Math Help - Definite Integral close??

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    76

    Definite Integral close??

    Find the derivative of

    h(x) = integral(sin(x) at the top and -5 at the bottom) of
    cos(t^5) +t

    So... here's what I thought the answer should be, but I'm evidently missing something, again.

    My plan of attack is to insert sin(x) for t and then subtract the function with -5 inserted for t:

    cos(sin^5(x)) + sin(x) - (cos((-5)^5) - 5)

    I'm getting told this is incorrect. Any glaring mistakes? I've been doing this for several hours, so I may have gone mentally blind.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    h'(x) = \frac{d}{dx} \int_{-5} ^{\sin x}\left( \cos t^5 + t\right)dt

    Main thing: Chain rule

    So as a 'modified' version of the fundamental theorem of calculus: Let F(x) = \int_a^{u(x)}f(t) \ dt. Then F'(x) = f \bigg( u(x) \bigg) u'(x) .

    Imagine f(t) = \cos t^5  + t and u(x) = \sin x.

    Apply the fundamental theorem of calculus as you would normally to get: h'(x) = \underbrace{\left[\cos \left(\sin^5 x\right) + \sin x\right]}_{f \big( u(x) \big)} \cdot \underbrace{\left(\sin x\right)'}_{u'(x)}

    Take the derivative of \sin x and you're done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    76
    So why didn't I have to take the -5 into consideration?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Well look at the fundamental theorem of calculus: \frac{d}{dx} \int_{a}^{u(x)} f(t) \ dt = f \bigg( u(x) \bigg) u'(x) ... or ... \frac{d}{dx} \int_{a}^{x} f(t) \ dt = f (x)

    In either case, you don't see the bottom limit of the integral, a, in the right hand side of the equation at all.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    Posts
    76
    Okay, I see what you mean, but in earlier problems I've been finding the answer by plugging in the upper limit and subtracting the lower limit. This had been giving me the correct answer until this question.

    For example, and earlier question said to find the derivative of upper limit = 1 lower = x
    sin(t^3) The correct answer ended up being sin(1^3) - sin(x^3) or just -sin(x)

    If I had followed the FTC and just used the upper limit I would've just gotten sin(1^3) or 0. How do I know when to do which?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    FTC says: \frac{d}{dx} \int_{a}^{x} f(t) \ dt = f (x) with x as the upper limit.

    If it is on the bottom, you'll have to make the adjustment by using the fact that: \int_a^b f(x) dx = -\int_b^a f(x) dx

    So for your question, by FTC: \frac{d}{dx} \int_{x}^{1} \sin t^3 \ dt = -\frac{d}{dx} \int_1^x \sin t^3 \ dt = -\sin x^3
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2008
    Posts
    76
    Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: December 5th 2011, 06:21 PM
  2. Replies: 4
    Last Post: April 13th 2011, 03:08 AM
  3. definite integral/ limit of integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2010, 05:00 AM
  4. Definite Integral Help!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 05:40 PM
  5. Definite integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2008, 09:37 PM

Search Tags


/mathhelpforum @mathhelpforum