Consider the following. x = t2 − 2t, y = t5, 1 ≤ t ≤ 4 Set up an integral that represents the length of the curve. 4 1 dt Use your calculator to find the length correct to four decimal places.

The length of the curve is 1023.0562 and this can be determined by doing the integration using the calculator.

Given :

First, differentiate x and y with respect to 't'.

Now, determine the length of the curve using the below formula:

Now, substitute the value of the known terms in the above formula and then integrate it.

Now, simplify the above integration using the calculator.

L = 1023.0562

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L ≈ 1023.0562

Step-by-step explanation:

We are given;

x = t² - 2t

dx/dt = 2t - 2

Also, y = t^(5)

dy/dt = 5t⁴

The arc length formula is;

L = (α,β)∫√[(dx/dt)² + (dy/dt)²]dt

Where α and β are the boundary points. Thus, applying this to our question, we have;

L = (1,4)∫√[(2t - 2)² + (5t⁴)²]dt

L = (1,4)∫√[4t² - 8t + 4 + 25t^(8)]dt

L = (1,4)∫√[25t^(8) + 4t² - 8t + 4]dt

Using online integral calculator, we have;

L ≈ 1023.0562

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry