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How do you set up an integral from the length of the curve #y=1/x, 1<=x<=5#?

Answer 1

Emma Aldridge

#int_1^5sqrt(1+1/x^4)#

The length of the curve of #y# on #alt=xlt=b# is equal to:

#L=int_a^bsqrt(1+(dy/dx)^2)#

For #y=1/x#, we see that #y=x^-1# so #dy/dx=-x^-2=-1/x^2#. Then:

#L=int_1^5sqrt(1+(-1/x^2)^2)=int_1^5sqrt(1+1/x^4)#

Which can be plugged into a calculator for a result of #4.15145#.

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Answer 2

Emma Aldridge

To set up the integral for the length of the curve y = 1/x from x = 1 to x = 5, you use the formula for arc length:

L = ∫√(1 + (dy/dx)^2) dx

First, find dy/dx:

dy/dx = -1/x^2

Then, square it:

(dy/dx)^2 = 1/x^4

Next, add 1 to (dy/dx)^2:

1 + (dy/dx)^2 = 1 + 1/x^4

Now, take the square root:

√(1 + (dy/dx)^2) = √(1 + 1/x^4)

Now, integrate with respect to x from 1 to 5:

L = ∫(1 to 5) √(1 + 1/x^4) dx

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you set up an integral from the length of the curve #y=1/x, 1&lt;=x&lt;=5#?

How do you set up an integral from the length of the curve #y=1/x, 1<=x<=5#?