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    Hi,

    I need to evaluate: \int^2_0 \frac{1}{x^2+25}\ dx

    My solution:

    Let x = 5u so \frac{dx}{du} = 5

    changing the limits: when x=0, u=0 and when x=2,u=10

    \int^2_0 \frac{1}{x^2+25}\ dx=\int^2_0 \frac{1}{25u^2+25}\ dx = \int^10_0 \frac{5}{25u^2+25}\ du = \frac{1}{5}\int^10_0 \frac{1}{u^2+1}\ du

    (Sorry, I can't quite get the last 2 integrals to look right, the weird bit at the beginning of each integral is supposed to say the integral between 10 and 0 of (fraction) du )

    = [(1/5) tan^-1 (u)] between 0 and 10

    = (1/5) tan^-1 (10) - (1/5) tan^-1 (0)

    = 0.294 (3sf)

    but the answer should be 0.0761. I think the problem is to do with the new limits, ie. the 10 should be 10/25 = 2/5 but I'm not sure why. Any help would be much appreciated. Thanks
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    if x = 5u then when x = 2 surely u = 2/5?
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    (Original post by Laurenje)
    Hi,

    I need to evaluate: \int^2_0 \frac{1}{x^2+25}\ dx

    My solution:

    Let x = 5u so \frac{dx}{du} = 5

    changing the limits: when x=0, u=0 and when x=2,u=10

    \int^2_0 \frac{1}{x^2+25}\ dx=\int^2_0 \frac{1}{25u^2+25}\ dx = \int^10_0 \frac{5}{25u^2+25}\ du = \frac{1}{5}\int^10_0 \frac{1}{u^2+1}\ du

    (Sorry, I can't quite get the last 2 integrals to look right, the weird bit at the beginning of each integral is supposed to say the integral between 10 and 0 of (fraction) du )

    = [(1/5) tan^-1 (u)] between 0 and 10

    = (1/5) tan^-1 (10) - (1/5) tan^-1 (0)

    = 0.294 (3sf)

    but the answer should be 0.0761. I think the problem is to do with the new limits, ie. the 10 should be 10/25 = 2/5 but I'm not sure why. Any help would be much appreciated. Thanks
    x=5u so u=0.2x. The new limits are one fifth of the originals, not five times.
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    Oops!! That's so typical of me, I do the hard bit, but then fail at the basics. Thank you.
 
 
 
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