Find the exact length of the curve y 1 4 x2 1 2 lnx 1 x 2


Introduction

In this article, we will discuss finding the exact length of the curve y 1 4 x2 1 2 lnx 1 x 2. We will first discuss what this curve represent, and then we will go over the formula used to find its length. After that, we will apply the formula to a few specific examples.

The Exact Length of the Curve y=1/4x^2+1/2lnx+1 on the Interval [1,2]

In this section, we will be finding the exact length of the curve y=1/4x^2+1/2lnx+1 on the interval [1,2]. We will be using the formula for the length of a curve, which is given by the integral of the square root of the sum of the squares of the first and second derivatives of the curve.

Finding the Arc Length of the Curve y=1/4x^2+1/2lnx+1

We want to find the length of the curve y=1/4x^2+1/2lnx+1 on the interval [1,2]. We can do this by finding the arc length of the curve.

The formula for arc length is:

L = ∫a→bf(x)dx

First, we need to find the derivative of our function. We can do this using the power rule and the natural log rule:

y′= 1⋅4x−2 + 1⋅ 2⋅ ln x− 1⋅ x− ln x
= 1⋅4x−2 + 2⋅ ln x− ln x
= 4x−2 − ln x

Now, we can plug our derivatives back into our arc length formula:

L = ∫1→2f(x)dx
= ∫1→24x−2 − lnxdx

To solve this integral, we can use integration by parts. We will let u=4x−2 and dv=lnxdx. Then we have:

du=4dx and v=lnxcotxx+C

Plugging these back into our equation, we get:

L=∫1→24x−2dx − ∫1→2lnxcotxx+Cdx

Now we can solve each integral separately. For the first one, we can use the power rule:

4∫1→24x−6dx=(4⋅ (−6))⋅ ∫1→24x−6dx=(−24)∫1→24x−6dx=12(4×12)=144[12]=1728

For the second integral, we need to use integration by parts again. This time, we will let u=lnxcotxx and dv=cotxx+C. Then we have:

du=(cotxx)(dx)+ (cotxx)(dv)=(cotxx)(dx)+ (cotxx)(cotxx+C)=(cotxx)2+(cotxx)(cotxx+C)

Plugging these back into our equation, we get:

L=∫1→21728 Cot^3(X)+ Cot^3(X)+ C dx

We can now solve this final integral using trigonometric identities:

L=∫1→21728 sec^3(X)+ tan^3(X)+ C dx=(sec^3(X))∫1→21728 sec^3(X)+ C dx+(tan^3(X))∫1→21728 sec^3(X)+ C dx
=1728sec^4(X)+ tan^3(X)sec^3(X)+C
=1728tanh^4(X/2)+ 1728coth^4(X/2)-8[8]
=10368-8[8]
=10360

The Exact Length of the Curve y=1/4x^2+1/2lnx+1 on the Interval [1,2]

The curve y=1/4x^2+1/2lnx+1 on the interval [1,2] has the exact length of 2.718. To find this, we’ll need to use the arc length formula: L=∫a→b√1+(dy/dx)^2dx

For this particular curve, we have dy/dx=(1/2)(1/(4x^2))=(1/8x) . Therefore, our equation for the arc length becomes: L=∫a→b√1+(dy/dx)^2dx=∫10.5≈12.5√129+((1/8)x)^210.5≈12.5

After doing the calculation, we get an answer of 2.718, which is the exact length of the curve y=1/4x^2+1/2lnx+1 on the interval [1,2].

Conclusion

After completing the calculations, we can conclude that the length of the curve y=1/4x^2+1/2lnx+1 from x=2 to infinity is equal to 2.718 units.


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