Question

hi I need to get line integral:

Integral( (4x+y)dx + (x+4y)dy ) and the curve is y=x^4 from A(-1,1) To B(1,1) please help

Answer 1

hi :)
since, y= x^4
you can plug in x^4 instead of 'y' in ( (4x+y)dx + (x+4y)dy )
also, for dy, you need to differentiate y=x^4 to get dy
have you tried these things ?

Answer 2

I have parametrized y and x like t=x so y=t*4 then I plugged it into the function and get incorrect answer the correct answer is -2, could you please write how you did it?
because when I try to solve it i think i am making everything fine but getting wrong answer thank you

Answer 3

ok, i get +2...maybe you could find any sign mistake in my work...
x=t --> dx=dt, y=t^4---->dy=4t^3dt
\((4t+x^4)dt +(t+4t^4)(4t^3)dt = (4t+x^4+4t^4+16t^7)dt\)
\(\large \int \limits_{-1}^1 (4t+x^4+4t^4+16t^7)dt = 0+ \int \limits_{-1}^1 5t^4 dt\)
because 't' and t^7 are odd functions of 'x' and when you integrate an odd function from -a to a , you get a 0
\(\large = 5 (t^5/5) = [t^5]_{-1}^1 = 1^5-(-1)^5=1-(-1)=2\)

Answer 4

that should be 4t+t^4 everywhere .....typing error

Answer 5

thats how t^4+4t^4 becomes 5t^4

Answer 6

don't think you need a separate parameter here, since you know y as a function of x you can substitute for y and make sure you take the differential of y w.r.t x

Answer 7

^correct, thats what i said in my first post, but using parameter is also not wrong, it'll yield same answer...

Answer 8

thanks lot

Answer 9

welcome ^_^

Answer 10

sorry, didn't read that :)