Evaluate the line integral, where c is the given curve. c y3 ds,
c. x = t3, y = t, 0 ≤ t ≤ 4

∫C) y³ × ds = 7962624,02 surface units

∫C) y³ × ds .               where  C    x = t³     y = t      with    0 ≤ t ≤ 4

∫C) y³ × ds   = ∫∫R) F(r(t) × dr

Then  F (r(t)) =   y³ . = t³ .          r ( t³ , t )  then .    dr/dt = (3×t² , 1 )

|| dr/dt|| = √ ( 3×t²)² + (1)²

|| dr/dt|| = √9×t⁴ + 1

∫C) y³ × ds   = ∫∫R) F(r(t) × dr . =  ∫₀⁴ t³ ×√9×t⁴ + 1    ×dt

To solve the integral we make a change of variables, we call

9×t⁴ + 1  = v .  then .    36×t³ ×dt = dv

Now as we change variables we need to change the limits of integration as follows:If   9×t⁴ + 1 = v .       then when t = 0   v = 1

And when  t = 4 .                then . v = 9×(4)⁴ + 1 . = 2305

The integral takes that form

∫₁²³⁰⁵ [ 1/36)× √v × dv . =   (1/36) ×∫₁²³⁰⁵ (v)¹/² dv  

= (1/36)×(2/3) × v³/² |₁²³⁰⁵ = (2/108)× √v³|₁²³⁰⁵ =

= (2/108) [ √( 9×t⁴ + 1 )³|₀⁴ =  (2/108) [ √( 9⁴×4¹⁶ + 1  - √(9)⁴×0 + 1)

=  (2/108) [√( (9⁴×4¹⁶ + 1 ) . + 1 ]

Neglecting  in √( (9⁴×4¹⁶ + 1 )  1 unit .  we get 9²×4⁸ = 429981696

= (2/108) [429981697]

= 7962624,02

250

step-by-step explanation:

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8