Evaluate the line integral c f · dr, where c is given by the vector function r(t). f(x, y) = xy i + 6y2 j r(t) = 14t4 i + t6 j, 0 ≤ t ≤ 1

Applying each separate step, it is found that the result of the line integral along the vector field is given by 58.

The curve is:

The vector field is:

Applying the vector field at the curve, we have that:

The derivative of the vector field is:

The dot product of the vector field along the curve with the derivative is:

Hence, the line integral is:

The result of the line integral is of 58.

A similar problem is given at link

Then the line integral is

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

∈3,600

step-by-step explanation: