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Thread: Area enclosed between curves

  1. #1
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    Area enclosed between curves

    Find the area between the curves 𝑦 = −𝑥 ^2 + 2𝑥 and 𝑦 = 𝑥 − 2 on the interval [0,3].

    I seem to be very poor at these in particular ...any advice on how to better them would be helpful

    First I set -x^2+2x=x-2
    I then did this -X^2+x+2=0
    because of the - sign in front of the x^2 I'm having trouble breaking it down ...how ever I wolfram alfa 'd it and the x values become -1 and 2

    As it is only looking for the interval between [0 and 3 ]

    I then integrated between -1 and 0 giving -13/6
    i then integrated between 3 and 0 and got 15/2
    or should I integrate between 2 and o giving 2/3

    the answer is 31/6 but I can't seem to figure what I must do ....I will practice these types ...hmmm
    feel free to unleash your guidance on my faults ...
    cheers ...
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  2. #2
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    Re: Area enclosed between curves

    Quote Originally Posted by bee77 View Post
    Find the area between the curves 𝑦 = −𝑥 ^2 + 2𝑥 and 𝑦 = 𝑥 − 2 on the interval [0,3].

    I seem to be very poor at these in particular ...any advice on how to better them would be helpful

    First I set -x^2+2x=x-2

    I then did this -X^2+x+2=0
    because of the - sign in front of the x^2 I'm having trouble breaking it down ...how ever I wolfram alfa 'd it and the x values become -1 and 2
    If you are taking calculus, you surely learned to solve quadratic equations long ago. That is the same as (multiply both sides by -1) x^2- x- 2= 0 (it isn't necessary to have the coefficient of x^2 positive but I prefer it that way).

    Now you can factor: since -2= (1)(-2) and -2+ 1= -1 x^2- x- 2= (x+ 1)(x- 2)= 0 so either x+ 1= 0 and x= 1 or x- 2= 0 and x= 2.

    Or you can "complete the square": x^2- x+ \frac{1}{4}- \frac{1}{4}- 2= \left(x- \frac{1}{2}\right)^2- \frac{9}{4}= 0 so \left(x- \frac{1}{2}\right)^2= \frac{9}{4}, x- \frac{1}{2}= \pm\frac{3}{2} and then x= \frac{1}{2}+ \frac{3}{2}= 2 or [tex]x= \frac{1}{2}- \frac{3}{2}= -1[tex].

    Or you can use the quadratic formula: x= \frac{-(-1)\pm\sqrt{(-1)^2- 4(1)(-2)}}{2}= \frac{1\pm\sqrt{9}}{2}= \frac{1\pm 3}{2} so x= \frac{4}{2}= 2 or x= \frac{-2}{2}= -1.

    As it is only looking for the interval between [0 and 3 ]

    I then integrated between -1 and 0 giving -13/6
    Why? As you say above "it is only looking for the interval between [0 and 3 ]" so what happens between -1 and 0 is irrelevant. The only reason to check for the zeroes of the function is to observe that the parabola, y= -x^2+ 2x is above the line y= x- 2 for x between 0 and 2 and below it for x between 2 and 3. You can either integrate (-x^2+ 2x)- (x- 2) from 0 to 2 and then integrate (x- 2)- (-x^2+ 2x) from 2 to 3 or (probably simpler) integrate (-x^2+ 2x)- (x- 2) from 0 to 2 and then subtract the integral of (-x^2+ 2x)- (x- 2) from 2 to 3.

    i then integrated between 3 and 0 and got 15/2
    or should I integrate between 2 and o giving 2/3

    the answer is 31/6 but I can't seem to figure what I must do ....I will practice these types ...hmmm
    feel free to unleash your guidance on my faults ...
    cheers ...
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

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    Re: Area enclosed between curves

    Quote Originally Posted by bee77 View Post
    Find the area between the curves �� = −�� ^2 + 2�� and �� = �� − 2 on the interval [0,3].

    I seem to be very poor at these in particular ...any advice on how to better them would be helpful

    First I set -x^2+2x=x-2

    I then did this -X^2+x+2=0
    because of the - sign in front of the x^2 I'm having trouble breaking it down ...how ever I wolfram alfa 'd it and the x values become -1 and 2
    If you are taking calculus, you surely learned to solve quadratic equations long ago. That is the same as (multiply both sides by -1) x^2- x- 2= 0 (it isn't necessary to have the coefficient of x^2 positive but I prefer it that way).

    Now you can factor: since -2= (1)(-2) and -2+ 1= -1 x^2- x- 2= (x+ 1)(x- 2)= 0 so either x+ 1= 0 and x= 1 or x- 2= 0 and x= 2.

    Or you can "complete the square": x^2- x+ \frac{1}{4}- \frac{1}{4}- 2= \left(x- \frac{1}{2}\right)^2- \frac{9}{4}= 0 so \left(x- \frac{1}{2}\right)^2= \frac{9}{4}, x- \frac{1}{2}= \pm\frac{3}{2} and then x= \frac{1}{2}+ \frac{3}{2}= 2 or x= \frac{1}{2}- \frac{3}{2}= -1.

    Or you can use the quadratic formula: x= \frac{-(-1)\pm\sqrt{(-1)^2- 4(1)(-2)}}{2}= \frac{1\pm\sqrt{9}}{2}= \frac{1\pm 3}{2} so x= \frac{4}{2}= 2 or x= \frac{-2}{2}= -1.

    As it is only looking for the interval between [0 and 3 ]

    I then integrated between -1 and 0 giving -13/6
    Why? As you say above "it is only looking for the interval between [0 and 3 ]" so what happens between -1 and 0 is irrelevant. The only reason to check for the zeroes of the function is to observe that the parabola, y= -x^2+ 2x is above the line y= x- 2 for x between 0 and 2 and below it for x between 2 and 3. You can either integrate (-x^2+ 2x)- (x- 2) from 0 to 2 and then integrate (x- 2)- (-x^2+ 2x) from 2 to 3 or (probably simpler) integrate (-x^2+ 2x)- (x- 2) from 0 to 2 and then subtract the integral of (-x^2+ 2x)- (x- 2) from 2 to 3.

    i then integrated between 3 and 0 and got 15/2
    or should I integrate between 2 and o giving 2/3

    the answer is 31/6 but I can't seem to figure what I must do ....I will practice these types ...hmmm
    feel free to unleash your guidance on my faults ...
    cheers ...
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  4. #4
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    Re: Area enclosed between curves

    wow thanks , yes I've done quadratic equations ...I will carefully go through the problem again and your notes ..the integration between 0 and 2 and then from 2 to 3 is something I'll work on ...So basically when you do find the x values working between the found values ie -1 to 2 but in this case 0 and 2 for the interval [0 and 3] and then 2 and 3 as it was one of the x values within the [0 and 3] interval as well is what I need to do in future similar problems (then subtracting the integrated values for area ) I think i got a bit confused as 3 wasn't one of the x values found but now it makes sense ....hope that makes sense ...thanks a million !!
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