LINE INTEGRAL (calcIII)


INTEGRAL_C (x − y) ds, where C is the portion of y = x^3 from (1, 1) to (−1, −1)

The integral is along C, so x - y = x - x³

ds = √(1 + (dy/dx)²) = √(1 + 9x^4)

The line integral becomes

 -1
 ∫ (x - x³)√(1 + 9x^4) dx
1

Break this up into two integrals

 -1
 ∫ x √(1 + 9x^4) dx -
1

 -1
 ∫ x³ √(1 + 9x^4) dx
1

For the first integral, use the substitution u = 3x²; for the second integral, use u = 1 + 9x^4