Question

这个高数题怎么求

Answer 1

y'=x²/2 - 1/2x²=(x^4 -1)/2x²
弧长s=∫[1,3]√(1+y'²)dx
=∫[1,3]√[1+(x^4 -1)²/ 4x^4]
=∫[1,3](x^4 +1)/2x² dx
=½∫[1,3](x²+ 1/x²)dx
=½ (x^3 /3 - 1/x)|[1,3]
=½(9- 1/3 - 1/3 +1)
=14/3