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Math Help - Separating Into Two Definite Integrals

  1. #1
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    Separating Into Two Definite Integrals

    I need to know the value of a and u for both integrals. I can then finish on my own.

    I found a to be 2(sqrt{x}) and u to be x but du/dx does not do away with the left side numerator which is 2x.

    See picture.
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  2. #2
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    Re: Separating Into Two Definite Integrals

    Let $u = 4x-x^2$. Then $du = (4-2x)dx$

    So,

    $\displaystyle \begin{align*}\int_2^3 \dfrac{2x-3}{\sqrt{4x-x^2}}dx & = -\int_2^3 \dfrac{3-2x}{\sqrt{4x-x^2}}dx \\ & = -\int_2^3 \dfrac{4-2x -1}{\sqrt{4x-x^2}}dx \\ & = -\int_2^3 \dfrac{4-2x}{\sqrt{4x-x^2}}dx + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = -\int_{u(2)}^{u(3)} u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx\end{align*}$

    Now, this second equation needs to have its square completed. The first integral does not.
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    Re: Separating Into Two Definite Integrals

    How does 2x-3 become 3-2x?
    You lost me there.
    Why did you replace 3-2x with 4-2x-1?
    What do you mean by setting the limits of integration to be u(2) and u(3)? Sorry but I don't follow the logic here. Can you explain what you did step by step?
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  4. #4
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    Re: Separating Into Two Definite Integrals

    $2x-3 = -(3-2x)$
    Note the negative sign outside the integral.

    As for setting the bounds of integration to $u(2)$ and $u(3)$, that is how integration works.

    $\displaystyle \int_a^b f(u(x))dx = \int_{u(a)}^{u(b)} f(u)du$
    This is one of the fundamental principles of the definite integral.

    For the $3-2x = 4-2x-1$, I was trying to get $du$ in the numerator. Since $du=(4-2x)dx$, I needed to turn $3-2x$ into $4-2x$. I did that by adding 1, then subtracting it.
    Last edited by SlipEternal; March 11th 2014 at 08:07 PM.
    Thanks from nycmath
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    Re: Separating Into Two Definite Integrals

    I understand but this integral is a bit tricky. Can you work it out completely? I can then use your reply as a guide to solve similar questions in my textbook.
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    Re: Separating Into Two Definite Integrals

    Ok, so, as above:

    $u = 4x-x^2, du = (4-2x)dx$

    $u(2) = 4(2) - (2)^2 = 4$
    $u(3) = 4(3) - (3)^2 = 3$

    So,

    $\displaystyle \begin{align*}\int_2^3 \dfrac{2x-3}{\sqrt{4x-x^2}}dx & = -\int_{u(2)}^{u(3)}u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = -\int_4^3 u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = \left. -2\sqrt{u} \right]_4^3 + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \end{align*}$

    Now, to complete the square for the second integral, you have

    $\displaystyle \begin{align*}4x-x^2 & = -(x^2-4x) \\ & = -\left(x^2-4x + \left(\frac{4}{2} \right)^2 - \left(\frac{4}{2} \right)^2\right) \\ & = -\left(x^2-4x+4-4\right) \\ & = -\left((x-2)^2-4\right) \\ & = 4-(x-2)^2\end{align*}$

    So, you have:

    $\displaystyle \begin{align*}(-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4-(x-2)^2}}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4\left(1-\left(\dfrac{x-2}{2}\right)^2 \right) }}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{1-\left(\dfrac{x-2}{2} \right)^2}}\left(\dfrac{1}{2} \right)dx\end{align*}$

    The final equality comes from taking the square root of 4 in the denominator and using commutativity and associativity to move it next to the $dx$.

    So, let $u = \dfrac{x-2}{2}$. Then $du = \dfrac{1}{2}dx$. This means $u(2) = 0$ and $u(3) = \dfrac{1}{2}$. So, you have:

    $\displaystyle \begin{align*}(-2\sqrt{3} + 4) + \int_2^3 \dfrac{1}{\sqrt{1-\left( \dfrac{x-2}{2} \right)^2}} \left(\dfrac{1}{2} \right)dx & = (-2\sqrt{3}+4) + \int_{u(2)}^{u(3)} \dfrac{1}{\sqrt{1-u^2}}du \\ & = (-2\sqrt{3}+4) + \left. \arcsin(u) \right]_0^{1/2} \\ & = (-2\sqrt{3}+4) + \left(\dfrac{\pi}{6}-0\right) \\ & = 4+\dfrac{\pi}{6}-2\sqrt{3}\end{align*}$

    Here it is through WolframAlpha (link). Wolframalpha does the same problem with trigonometric substitution which makes it much easier to solve.
    Thanks from nycmath
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    Re: Separating Into Two Definite Integrals

    Quote Originally Posted by SlipEternal View Post
    Ok, so, as above:

    $u = 4x-x^2, du = (4-2x)dx$

    $u(2) = 4(2) - (2)^2 = 4$
    $u(3) = 4(3) - (3)^2 = 3$

    So,

    $\displaystyle \begin{align*}\int_2^3 \dfrac{2x-3}{\sqrt{4x-x^2}}dx & = -\int_{u(2)}^{u(3)}u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = -\int_4^3 u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = \left. -2\sqrt{u} \right]_4^3 + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \end{align*}$

    Now, to complete the square for the second integral, you have

    $\displaystyle \begin{align*}4x-x^2 & = -(x^2-4x) \\ & = -\left(x^2-4x + \left(\frac{4}{2} \right)^2 - \left(\frac{4}{2} \right)^2\right) \\ & = -\left(x^2-4x+4-4\right) \\ & = -\left((x-2)^2-4\right) \\ & = 4-(x-2)^2\end{align*}$

    So, you have:

    $\displaystyle \begin{align*}(-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4-(x-2)^2}}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4\left(1-\left(\dfrac{x-2}{2}\right)^2 \right) }}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{1-\left(\dfrac{x-2}{2} \right)^2}}\left(\dfrac{1}{2} \right)dx\end{align*}$

    The final equality comes from taking the square root of 4 in the denominator and using commutativity and associativity to move it next to the $dx$.

    So, let $u = \dfrac{x-2}{2}$. Then $du = \dfrac{1}{2}dx$. This means $u(2) = 0$ and $u(3) = \dfrac{1}{2}$. So, you have:

    $\displaystyle \begin{align*}(-2\sqrt{3} + 4) + \int_2^3 \dfrac{1}{\sqrt{1-\left( \dfrac{x-2}{2} \right)^2}} \left(\dfrac{1}{2} \right)dx & = (-2\sqrt{3}+4) + \int_{u(2)}^{u(3)} \dfrac{1}{\sqrt{1-u^2}}du \\ & = (-2\sqrt{3}+4) + \left. \arcsin(u) \right]_0^{1/2} \\ & = (-2\sqrt{3}+4) + \left(\dfrac{\pi}{6}-0\right) \\ & = 4+\dfrac{\pi}{6}-2\sqrt{3}\end{align*}$

    Here it is through WolframAlpha (link). Wolframalpha does the same problem with trigonometric substitution which makes it much easier to solve.
    Thank you very much. I could have never done all that work without getting lost. I hope this textbook gets easier.
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  8. #8
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    Re: Separating Into Two Definite Integrals

    This is a calculus 1 integral. I would hate to see what a calculus 2 and 3 integral looks like.
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