Titik yang dilalui oleh y=\dfrac4{x^2} untuk x=1
{y=\dfrac4{1^2}=4}
(1,4)
\(\eqalign{
c&=\dfrac4{x^2}\\
x^2&=\dfrac4c\\
x&=\dfrac2{\sqrt{c}}
}\)
\(\eqalign{
L&=2\dfrac14\\
\displaystyle\intop_1^{\frac2{\sqrt{c}}}\left(\dfrac4{x^2}-c\right)\ dx&=\dfrac94\\
\displaystyle\intop_1^{\frac2{\sqrt{c}}}\left(4x^{-2}-c\right)\ dx&=\dfrac94\\
\left.\dfrac4{-1}x^{-1}-cx\right|_1^{\frac2{\sqrt{c}}}&=\dfrac94\\
\left.-\dfrac4x-cx\right|_1^{\frac2{\sqrt{c}}}&=\dfrac94\\
\left(-\dfrac4{\dfrac2{\sqrt{c}}}-c\left(\dfrac2{\sqrt{c}}\right)\right)-\left(-\dfrac41-c(1)\right)&=\dfrac94\\
\left(-2\sqrt{c}-2\sqrt{c}\right)-(-4-c)&=\dfrac94\\
-4\sqrt{c}+4+c&=\dfrac94\\
c-4\sqrt{c}+\dfrac74&=0\\
4c-16\sqrt{c}+7&=0\\
\left(2\sqrt{c}-1\right)\left(2\sqrt{c}-7\right)&=0
}\)
\(\eqalign{
2\sqrt{c_1}-1&=0\\
2\sqrt{c_1}&=1\\
\sqrt{c_1}&=\dfrac12\\
c_1&=\dfrac14
}\) | \(\eqalign{
2\sqrt{c_2}-7&=0\\
2\sqrt{c_2}&=7\\
\sqrt{c_2}&=\dfrac72\\
c_2&=\dfrac{49}4
}\) |
c_1+c_2=\dfrac14+\dfrac{49}4=\dfrac{50}4=\boxed{\boxed{\dfrac{25}2}}
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