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Math Help - Applications of Integration

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    is up to his old tricks again! Jhevon's Avatar
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    Applications of Integration

    I am posting this on behalf of vc15ao4 who seems to have problems posting.

    Quote Originally Posted by vc15ao4
    I can't seem to figure out how to post in the forum. Every time I'm at the forum, it says I can't. I dont have enough access privledges it says?? Maybe you can help me with that, but i do have a few more problems i need help on

    1) Evaluate the integral: (long looking s?) e^(ax+b)dx

    2) Use a graphing utility to graph f(x)= 4-x/x. Then determine the area of the region in the first quadrant bounded by f, the xaxis, and x=1.

    3) Use a graphing utility as an aid to sketch the region bounded by the fuction f(x)= 1/(square roof of 9-x^2) , the x axis, and the lines x=1, and x=2. Then calculate the area of the region
    1) Evaluate the integral: (long looking s?) e^(ax+b)dx
    i, and a lot of other people on this site, write the "long looking s" as "int" for integral--that is if we're not using LaTex.

    there's a shortcut to find the integral of a function like this, that is a function where we have e raised to some polynomial in x with the highest power of x being 1, but its probably dangerous telling you that at this point, so let's do this the proper way, by substitution.

    int{e^(ax+b)}dx
    let u = ax + b
    => du = a dx ...........took the derivative of ax + b with respect to x
    => 1/a du = dx

    so our integral becomes:

    (1/a)*int{e^u}du
    = (1/a)e^u + C
    = (1/a)e^(ax + b) + C




    2) Use a graphing utility to graph f(x)= 4-x/x. Then determine the area of the region in the first quadrant bounded by f, the xaxis, and x=1.
    see the diagram below called areaquestion2, the shaded region is the region we want the area of.

    we know that one of the limits of integration is x = 1, but what's the other? we need to find where (4 - x)/x cuts the x-axis. in other words, we want the point(s) where

    y = (4 - x)/x intersects y = 0

    => (4 - x)/x = 0
    => 4 - x = 0
    => x = 4

    so our limits of integration are 1 to 4

    so A = int{(4 - x)/x}dx evaluated between 1 and 4
    int{(4 - x)/x}dx
    = int{4/x - 1}dx
    = 4ln(x) - x + C evaluated between 1 and 4

    => A = 4ln(4) - 4 - 4ln(1) + 1
    => A = 4ln(4) - 3 ................ln(1) = 0


    3) Use a graphing utility as an aid to sketch the region bounded by the fuction f(x)= 1/(square roof of 9-x^2) , the x axis, and the lines x=1, and x=2. Then calculate the area of the region
    i'm almost certain i've done this exact problem on this site before...o well

    by the way, it's "squareroot" not "square roof," but i guess that's an honest mistake

    see the diagram below called areaquestion3, the shaded region is the part we want the area of

    our limits of integration are obviously x = 1...2, so now

    A = int{1/sqrt(9 - x^2)}dx

    let's simplify this a bit, factor out the 9 out of the bottom of the squareroot, you'll see why

    => A = int{1/sqrt[9(1 - (x^2)/9]}dx ...........simplify a bit more
    => A = int{1/sqrt[9(1 - (x/3)^2)]}dx
    => A = int{1/[sqrt(9) * sqrt(1 - (x/3)^2)}dx ........you should be able to see it now, i was going for arcsine
    => A = (1/3)*int{1/sqrt(1 - (x/3)^2)}dx ..........i changed 1/sqrt(9) to 1/3 and factored it out

    again, there's a shortcut here, but let's be proper

    now let u = x/3
    => du = 1/3 dx
    => 3 du = dx
    so our integral becomes:

    (1/3)(3)int{1/sqrt(1 - u^2)}du

    = arcsin(u) + C
    = arcsin(x/3) + C .........now we evaluate between our limits

    => A = arcsin(2/3) - arcsin(1/3)
    Attached Thumbnails Attached Thumbnails Applications of Integration-areaquestion2.gif   Applications of Integration-areaquestion3.gif  
    Last edited by Jhevon; March 25th 2007 at 07:05 PM.
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  2. #2
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    Quote Originally Posted by Jhevon View Post

    int{e^(ax+b)}dx
    let u = ax + b
    => du = a dx ...........took the derivative of ax + b with respect to x
    => 1/a du = dx
    If a!=0
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If a!=0
    correct.

    if a = 0

    then int{e^(ax+b)}dx = int{e^b}dx = xe^b + C ........since e^b is a constant
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