3. (25 points) find the arc length of the curve \n$y = \\frac{(x + 1)^3}{3} + \\frac{1}{4x + 4}$\nbetween $x…

Explanation:

Step1: Recall the arc length formula

The arc length ( L ) of a curve ( y = f(x) ) from ( x = a ) to ( x = b ) is given by ( L=\int_{a}^{b}\sqrt{1+(f^{\prime}(x))^{2}}dx ). First, we need to find the derivative ( y^{\prime} ) of the given function.

Step2: Find the derivative ( y^{\prime} )

Given ( y=\frac{(x + 1)^{3}}{3}+\frac{1}{4(x + 1)} ) (rewriting ( 4x + 4=4(x + 1) ) for easier differentiation). Using the power rule and the chain rule:

• The derivative of ( \frac{(x + 1)^{3}}{3} ) is ( \frac{3(x + 1)^{2}}{3}=(x + 1)^{2} ).
• The derivative of ( \frac{1}{4(x + 1)}=\frac{1}{4}(x + 1)^{-1} ) is ( \frac{1}{4}\times(-1)(x + 1)^{-2}=-\frac{1}{4(x + 1)^{2}} ).

So, ( y^{\prime}=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}} ).

Step3: Compute ( 1+(y^{\prime})^{2} )

Let's square ( y^{\prime} ): ( (y^{\prime})^{2}=\left[(x + 1)^{2}-\frac{1}{4(x + 1)^{2}}\right]^{2}=(x + 1)^{4}-2\times(x + 1)^{2}\times\frac{1}{4(x + 1)^{2}}+\frac{1}{16(x + 1)^{4}} ) Simplify the middle term: ( -2\times\frac{1}{4}=-\frac{1}{2} ) So, ( (y^{\prime})^{2}=(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}} ) Then, ( 1+(y^{\prime})^{2}=1+(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}}=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}} )

Notice that ( (x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}}=\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]^{2} ). Let's verify: ( \left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]^{2}=(x + 1)^{4}+2\times(x + 1)^{2}\times\frac{1}{4(x + 1)^{2}}+\frac{1}{16(x + 1)^{4}}=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}} ), which matches ( 1+(y^{\prime})^{2} ).

Step4: Simplify the square root

Since ( \sqrt{1+(y^{\prime})^{2}}=\sqrt{\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]^{2}} ), and for ( x\in[0,1] ), ( (x + 1)^{2}>0 ) and ( \frac{1}{4(x + 1)^{2}}>0 ), so the square root is ( (x + 1)^{2}+\frac{1}{4(x + 1)^{2}} ).

Step5: Compute the integral

Now, we need to compute ( L=\int_{0}^{1}\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]dx ) Split the integral: ( L=\int_{0}^{1}(x + 1)^{2}dx+\frac{1}{4}\int_{0}^{1}(x + 1)^{-2}dx )

For the first integral, let ( u=x + 1 ), ( du=dx ). When ( x = 0 ), ( u = 1 ); when ( x = 1 ), ( u = 2 ). So ( \int_{1}^{2}u^{2}du=\left[\frac{u^{3}}{3}\right]_{1}^{2}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3} ).

For the second integral, let ( v=x + 1 ), ( dv=dx ). When ( x = 0 ), ( v = 1 ); when ( x = 1 ), ( v = 2 ). So ( \frac{1}{4}\int_{1}^{2}v^{-2}dv=\frac{1}{4}\left[\frac{v^{-1}}{-1}\right]_{1}^{2}=\frac{1}{4}\left(-\frac{1}{2}+1\right)=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8} ).

Now, add the two results: ( L=\frac{7}{3}+\frac{1}{8}=\frac{56 + 3}{24}=\frac{59}{24} )? Wait, no, wait: Wait, ( \frac{7}{3}+\frac{1}{8}=\frac{56+3}{24}=\frac{59}{24} )? Wait, no, let's recalculate the second integral:

Wait, ( \int v^{-2}dv=\frac{v^{-1}}{-1}+C=-\frac{1}{v}+C ). So ( \frac{1}{4}\left[-\frac{1}{v}\right]{1}^{2}=\frac{1}{4}\left(-\frac{1}{2}+\frac{1}{1}\right)=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8} ). And the first integral: ( \int{1}^{2}u^{2}du=\left[\frac{u^{3}}{3}\right]_{1}^{2}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3} ). Then ( \frac{7}{3}+\frac{1}{8}=\frac{56 + 3}{24}=\frac{59}{24} )? Wait, no, that can't be. Wait, wait, I think I made a mistake in the sign when squaring. Wait, ( y^{\prime}=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}} ), then ( (y^{\prime})^{2}=(x + 1)^{4}-2\times(x + 1)^{2}\times\frac{1}{4(x + 1)^{2}}+\frac{1}{16(x + 1)^{4}}=(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}} ). Then ( 1+(y^{\prime})^{2}=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}}=\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]^{2} ), that's correct. Then the square root is ( (x + 1)^{2}+\frac{1}{4(x + 1)^{2}} ), correct.

Wait, but let's re - compute the integral:

First integral: ( \int_{0}^{1}(x + 1)^{2}dx=\left[\frac{(x + 1)^{3}}{3}\right]_{0}^{1}=\frac{(2)^{3}}{3}-\frac{(1)^{3}}{3}=\frac{8 - 1}{3}=\frac{7}{3} ).

Second integral: ( \frac{1}{4}\int_{0}^{1}(x + 1)^{-2}dx=\frac{1}{4}\left[\frac{(x + 1)^{-1}}{-1}\right]_{0}^{1}=\frac{1}{4}\left(-\frac{1}{2}+1\right)=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8} ).

Wait, but ( \frac{7}{3}+\frac{1}{8}=\frac{56 + 3}{24}=\frac{59}{24}\approx2.458 ). But let's check the derivative again. Wait, the original function is ( y=\frac{(x + 1)^{3}}{3}+\frac{1}{4x + 4}=\frac{(x + 1)^{3}}{3}+\frac{1}{4(x + 1)} ). The derivative of ( \frac{(x + 1)^{3}}{3} ) is ( (x + 1)^{2} ), correct. The derivative of ( \frac{1}{4(x + 1)} ) is ( -\frac{1}{4(x + 1)^{2}} ), correct. So ( y^{\prime}=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}} ), correct. Then ( (y^{\prime})^{2}=(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}} ), then ( 1+(y^{\prime})^{2}=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}}=\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]^{2} ), correct. So the square root is ( (x + 1)^{2}+\frac{1}{4(x + 1)^{2}} ), correct.

Wait, but let's compute the integral again:

( \int_{0}^{1}(x + 1)^{2}dx=\left[\frac{(x + 1)^3}{3}\right]_0^1=\frac{8}{3}-\frac{1}{3}=\frac{7}{3} )

( \int_{0}^{1}\frac{1}{4(x + 1)^2}dx=\frac{1}{4}\int_{0}^{1}(x + 1)^{-2}dx=\frac{1}{4}\left[-\frac{1}{x + 1}\right]_0^1=\frac{1}{4}\left(-\frac{1}{2}+1\right)=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8} )

Then ( L=\frac{7}{3}+\frac{1}{8}=\frac{56 + 3}{24}=\frac{59}{24} )? Wait, no, ( \frac{7}{3}=\frac{56}{24} ), ( \frac{1}{8}=\frac{3}{24} ), so ( \frac{56 + 3}{24}=\frac{59}{24} ). Wait, but let's check with another approach. Let's expand ( (x + 1)^{2}+\frac{1}{4(x + 1)^{2}}=x^{2}+2x + 1+\frac{1}{4(x^{2}+2x + 1)} ). But maybe there was a miscalculation in the square of ( y^{\prime} ). Wait, ( y^{\prime}=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}} ), so ( (y^{\prime})^{2}=(x + 1)^{4}-2\times(x + 1)^{2}\times\frac{1}{4(x + 1)^{2}}+\frac{1}{16(x + 1)^{4}}=(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}} ). Then ( 1+(y^{\prime})^{2}=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}} ), which is a perfect square: ( A^{2}+2AB + B^{2} ), where ( A=(x + 1)^{2} ), ( B=\frac{1}{4(x + 1)^{2}} ), since ( 2AB = 2\times(x + 1)^{2}\times\frac{1}{4(x + 1)^{2}}=\frac{1}{2} ), which matches the middle term. So the square root is correct. So the integral calculation should be correct.

Answer:

(\frac{59}{24})