# Evaluate ∫f(xy)ds f(xy)=(x^2)(y^3)dy if C is the arc of the curve from (-1,1) to (8,2)

*print*Print*list*Cite

### 1 Answer

You need to parametrize the curve C:

`x=(1-t)x_0+tx_1=gtx=(1-t)*(-1)+8t=gtx=t-1+8t = 9t-1`

`y=(1-t)y_0+ty_1=gtx=(1-t)*(1)+2t=gty=1-t+8t = 7t+1`

You need to find ds:

`ds= sqrt((dx/dt)^2 + (dy/dt)^2) = sqrt(81+49)=gt ds = sqrt 130`

```oint` c(x,y)ds =`int_0^1 sqrt 130*(9t-1)^2*(7t+1)^3dt`

Expand the binomials:

`(9t-1)^2 = 81t^2 - 18t + 1`

`(7t+1)^3 = 343t^3 + 147t^2 + 21t + 1`

`(9t-1)^2*(7t+1)^3 = (81t^2 - 18t + 1)(343t^3 + 147t^2 + 21t + 1)`

`(9t-1)^2*(7t+1)^3 = 27783t^5 + 11907t^4 + 1701t^3 + 81t^2 - 6147t^3 - 378t^2 - 18t + 343t^3 + 147t^2 + 21t + 1`

`int_0^1 sqrt 130*(9t-1)^2*(7t+1)^3dt = sqrt 130(27783/6 + 11907/5 + 81/3 - 6147/4 - 378/3 - 18/2 + 343/4 + 147/3 + 21/2 + 1)`

`int_0^1 sqrt 130*(9t-1)^2*(7t+1)^3dt = sqrt 130(4630.5 + 2381.4 + 27 - 1536.75 - 126 - 9 + 85.75 + 49 + 10.5 + 1)`

`int_0^1 sqrt 130*(9t-1)^2*(7t+1)^3dt = 5513.4sqrt 130`

**Evaluating the line integral yields `5513.4sqrt 130.` **