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Calc. II

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uklune | eNotes Newbie

Posted October 31, 2013 at 8:48 PM via web

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Calc. II

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted October 31, 2013 at 10:23 PM (Answer #1)

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For calculating the arc length of curve `y=f(x)` you  use the following formula

`s=int_a^bsqrt(1+y'^2)dx`

Now we first calculate `y'`

`y'=(e^x+2)/(e^x-2)cdot(e^x(e^x+2)-(e^x-2)e^x)/(e^x+2)^2=`

`(e^x(e^x-2-e^x+2))/(e^(2x)-4)=`

`y'=(4e^x)/(e^(2x)-4)`

Now we insert that into our formula.

`s=int_1^2sqrt(1+(16e^(2x))/(e^(4x)-8e^(2x)+16))dx=`

`int_1^2sqrt((e^(4x)+8e^(2x)+16)/(e^(4x)-8e^(2x)+16))dx=`

`int_1^2(e^(2x)+4)/(e^(2x)-4)dx=`

`int_1^2e^(2x)/(e^(2x)-4)dx+4int_1^2dx/(e^(2x)-4)`

Now we solve each integral separately.

`I_1=int_1^2e^(2x)/(e^(2x)-4)dx=|(t=e^(2x)-4),(dt=2e^(2x)dx),(t_1=e^2-4),(t_2=e^4-4)|=`

`1/2int_(e^2-4)^(e^4-4)dt/t=1/2(ln(e^4-4)-ln(e^2-4))=`

`1/2ln((e^4-4)/(e^2-4))`

`I_2=4int_1^2dx/(e^(2x)-4)=|(t=e^(2x)),(dt=2e^(2x)dx),(dx=dt/2t),(t_1=e^2),(t_2=e^4)|=`

`2int_(e^2)^(e^4)dt/((t-4)t)=`

Now we use partial fractions.

`2int_(e^2)^(e^4)(1/(4(t-4))-1/(4t))dt=` `1/2(ln(t-4)-lnt)|_(e^2)^(e^4)=` `1/2(ln(e^4-4)-4-ln(e^2-4)+2)=` `1/2ln((e^4-4)/(e^2-4))-1`  

Now to obtain final result we just add `I_1` and `I_2.`

`s=I_1+I_2=1/2ln((e^4-4)/(e^2-4))+1/2ln((e^4-4)/(e^2-4))-1=`

` `  `ln((e^4-4)/(e^2-4))-1=`

`s=ln((e^4-4)/(e^3-4e))`

The arc length of the curve is `ln((e^4-4)/(e^3-4e))`

Sketch of the curve is given in the image below.

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