y = x^5/6+1/{10x^3} impliesy=x56+110x3⇒
dy/dx= (5x^4)/6-3/(10x^4) impliesdydx=5x46−310x4⇒
1+(dy/dx)^2 = 1+( (5x^4)/6-3/(10x^4))^21+(dydx)2=1+(5x46−310x4)2
Now, since 1 = 4 times (5x^4)/6 times 3/(10x^4)1=4×5x46×310x4, the expression on the right above simplifies !
1+(dy/dx)^2 =( (5x^4)/6+3/(10x^4))^21+(dydx)2=(5x46+310x4)2
Thus, the required arc length
L = int_a^b sqrt{1+(dy/dx)^2}dx = int_1^2( (5x^4)/6+3/(10x^4)) dx = (x^5/6-1/(10x^3))_1^2 =(2^5/6-1/(10times 2^3))-(1/6-1/10) = 1219/240 L=∫ba√1+(dydx)2dx=∫21(5x46+310x4)dx=(x56−110x3)21=(256−110×23)−(16−110)=1219240
