**Help with evaluating the line integral ∫c y dx +x^2 dy?**

Someone on a different internet website helped me with this line integration problem but I can't see how they got some of the steps and my Calculus book doesn't explain it well: The answers are supposed to be: a) 100/3, b) 60

I did read the post "Read Before Posting". My problem is that I am missing some important steps for how to solve this problem, but I can't solve it if I don't know what to do first. I know about derivatives but maybe I am not using them correctly.

Evaluate the line integral ∫c y dx + x^2 dy a) along the curve x = 2t, y = t^2 - 1 from t = 0 to t = 2 b) from the point A(0, -1) to the point B (4, 3) along the line segments AC and CB, where C is the point (4, -1) Hint: Along AC, dy = 0, along CB, dx = 0

a) = [(t^2 - 1) * 2 + (2t)^2 * 2t] dt, since x = 2t and y = t^2 - 1 = (8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) {for t = 0 to 2} = 100/3

**I don't understand how they got the 8t^3 + 2t^2 – 2)dt and why it equals 2t^4 + (2/3)t^3 - 2t and why that equals 100/3:**

(8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) = 100/3

**Also: Does dx = 0? Because the derivative of 4 = 0?**

Does dy = 0? Because the derivative of -2 = 0?

What does the derivative of dt = ?? I don't know what this would be.

Does dy = 0? Because the derivative of -2 = 0?

What does the derivative of dt = ?? I don't know what this would be.

(Do this one segment at a time and add the results):

b) For AC, we have y = -1 and so dy = 0. ∫AC (y dx + x^2 dy) = ∫(x = 0 to 4) [-1 * 1 + x^2 * 0] dx = -4.

**Why and How does [-1 * 1 + x^2 * 0] dx = -4**? The person who helped me didn't explain how they got this and I can't figure it out.

For CB, we have x = 4 and so dx = 0. ∫CB (y dx + x^2 dy) = ∫(y = -1 to 3) (y * 0 + 4^2 * 1) dy = 64.

**Why and how does (y * 0 + 4^2 * 1) dy = 64?**The person who helped me did not explain how they got this either. For b) ∫c (y dx + x^2 dy) = -4 + 64 = 60.

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