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Math Help - Reversing the order of integration

  1. #1
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    Reversing the order of integration

    Question:
    Evaluate the iterated integral \int_0^1{\int_\frac{x}{2}^\frac{1}{2}{e^{y^2}}\,dy  }\,dx by reversing the order of integration.

    Is this correct?
    current limits are:
    \frac{x}{2} \leq y \leq \frac{1}{2} = x \leq 2y \leq 1 and 0 \leq x \leq 1
    rearranging these gives:
    0 \leq y \leq \frac{1}{2} and 0 \leq x \leq 2y

    making the equation now \int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy

    Correct? yes/no? Want some reassurance before I try to tackle the integration part!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Dr Zoidburg View Post
    Question:
    Evaluate the iterated integral \int_0^1{\int_\frac{x}{2}^\frac{1}{2}{e^{y^2}}\,dy  }\,dx by reversing the order of integration.

    Is this correct?
    current limits are:
    \frac{x}{2} \leq y \leq \frac{1}{2} = x \leq 2y \leq 1 and 0 \leq x \leq 1
    rearranging these gives:
    0 \leq y \leq \frac{1}{2} and 0 \leq x \leq 2y

    making the equation now \int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy

    Correct? yes/no? Want some reassurance before I try to tackle the integration part!
    This is correct. I always find it a good exercise to draw the domain (whenever possible) so that I can visually see what we're trying to do.
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  3. #3
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    cool. Always glad to know I'm on the right track!

    For the equation itself, is this correct:
    \int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy
    =
    \int_0^\frac{1}{2}{xe^{y^2}\bigg|_{x=0}^{x=2y}}\,d  y
    =
    \int_0^\frac{1}{2}{2y*e^{y^2}}\,dy

    setting u=y^2 gives du = 2ydy
    giving:
    \int_{u=0}^{u=\frac{1}{2}}e^u\,du

    =e^{\frac{1}{2}} - 1 = 0.6487 (4 dp)
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  4. #4
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    Quote Originally Posted by Dr Zoidburg View Post
    cool. Always glad to know I'm on the right track!

    For the equation itself, is this correct:
    \int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy
    =
    \int_0^\frac{1}{2}{xe^{y^2}\bigg|_{x=0}^{x=2y}}\,d  y
    =
    \int_0^\frac{1}{2}{2y*e^{y^2}}\,dy

    setting u=y^2 gives du = 2ydy
    giving:
    \int_{u=0}^{u=\frac{1}{2}}e^u\,du

    =e^{\frac{1}{2}} - 1 = 0.6487 (4 dp)
    Yes, that is correct- although I would have left the answer as e^{\frac{1}{2}}- 1 rather than write an approximation.
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