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Math Help - Monotonic functions?

  1. #1
    Super Member fardeen_gen's Avatar
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    Monotonic functions?

    Let a + b =4, a < 2 and g(x) be monotonically increasing function of x. Prove that f(x) = \int_{0}^{a} g(x)\ dx + \int_{0}^{b} g(x)\ dx increases with increase in (b - a).
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Let a + b =4, a < 2 and g(x) be monotonically increasing function of x. Prove that \color{red}f(x) = \int_{0}^{a} g(x)\ dx + \int_{0}^{b} g(x)\ dx increases with increase in (b - a).
    You need to repair this post.
    As written f(x) is constant.

    Also what does "increases with increase in" mean?
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  3. #3
    Super Member fardeen_gen's Avatar
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    I have no idea what the question means! Thats why I posted it.
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  4. #4
    Senior Member pankaj's Avatar
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    You are right Plato.

    The question says that prove that \int_{0}^{a}g(x)dx+\int_{0}^{b}g(x)dx increases as (b-a) increases.

    Let b-a=t.
    Therefore, a=\frac{4-t}{2}
     <br />
b=\frac{4+t}{2}<br />

    Let F(t)=\int_{0}^{a}g(x)dx+\int_{0}^{b}g(x)dx=\int_{0  }^{\frac{4-t}{2}} g(x)dx+\int_{0}^{\frac{4+t}{2}} g(x)dx

     <br />
F'(t)=g(\frac{4-t}{2})(-\frac{1}{2})+g(\frac{4+t}{2})(\frac{1}{2})<br />
    (Using Leibnitz rule)

     <br />
F'(t)=\frac{1}{2}(g(b)-g(a))<br />

    Given that a<2,therefore b>2,i.e.b>a.

    Now,since g(x) is monotonically increasing,therefore g(b)>g(a)

    Hence, F'(t)>0

    Therefore, F(t) increases as t increases

    i.e. \int_{0}^{a}g(x)dx+\int_{0}^{b}g(x)dx increases as (b-a) increases.
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