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Simple Integration and Differentiation Problem

  1. Jun 20, 2013 #1
    The question is the following:

    y = (lower limit 0, upper limit e2x)∫ (1/T)

    Find y'(2)

    According to the markscheme, the answer is equal to 4e2. I got 2e2.

    I did the following:

    I integrated the expression, which yielded: 2√t. I then substituted the upper and lower limits in, which gave me:

    2ex

    I then differentiated that, which yields:

    2ex

    I proceeded to substitute 2 in, which gave me 2e2

    Can anyone shed some light onto this for me please?

    Thank you in advance!
     
  2. jcsd
  3. Jun 20, 2013 #2
    Could u write it on a paper and post. I'm on phone, can't read the equation clearly. If I could I will try help
     
  4. Jun 20, 2013 #3

    Zondrina

    User Avatar
    Homework Helper

    Are you familiar with the fundamental theorem of calculus? If so, what is the result of this :

    $$\frac{d}{dx} y = \frac{d}{dx} \int_{0}^{e^{2x}} \frac{1}{t} dt$$
     
  5. Jun 20, 2013 #4
    I forgot to add (and this might answer your question, Zondrina) that, my first instinct was the following: If y is equal to the integral of 1/t, then the derivative of y should be equal to 1/t. But, if I went on and substituted 1/2 into the equation, I would be left with 1/2, which is not equal to any of the options available.

    Anyway, is the answer to what you asked me (1/T)? :blushing:
     
  6. Jun 20, 2013 #5

    Mark44

    Staff: Mentor

    You are really oversimplifying things. Here is what is true:
    $$ \text{If } f(x) = \int_0^x \frac{dt}{t}$$
    then f'(x) = 1/x

    The fact that your integral has an upper limit of e2x rather than x is very significant, and means that you will need to apply the chain rule when you differentiate.
    Also, do not delete the homework template when you post a problem. It's there for a reason.
     
  7. Jun 20, 2013 #6

    Zondrina

    User Avatar
    Homework Helper

    An easy way to remember the theorem is this :

    $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))(b'(x)) - f(a(x))(a'(x))$$

    It will give you the correct answer.
     
  8. Jun 20, 2013 #7

    Mark44

    Staff: Mentor

    Speaking only for myself, I find it simpler to use the chain rule directly than to remember a canned formula (that could possibly be remembered incorrectly).

    $$\frac{d}{dx} \int_a^{u(x)} f(t) dt =\frac{d}{du} \left( \int_a^{u(x)} f(t) dt \right) * \frac{du}{dx}$$
    $$=f(u(x)) * \frac{du}{dx}$$

    Applying this idea to the original integral entails splitting it into two integrals.
     
  9. Jun 20, 2013 #8
    Thank you very much to both of you. As a matter of fact, I knew the "canned formula" but had never been introduced to the concept of chain rule through this way, hence, I had no idea I had to apply it here.

    Thank you once again and sorry for deleting the template.
     
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