## 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.