Find the arc length for y=x^2+x.

Step-by-step solution-:

The arc length of function y=f(x) from a< x< b, is given by:

s(x)=\int_{a}^{b}\sqrt{1+(f'(x))^{2}}dx

Evaluate the first derivative:

\frac{d}{dx}{(x^2+x)}=2x+1

Hence, we have:

f'(x)=2x+1,  -1\leq{x}\leq0

s(x)=\int_{\textcolor{primary}{a}}^{\textcolor{secondary}{b}}\sqrt{1+(\textcolor{tertiary}{f'(x)})^{2}}~dx

Substitute a=-1, b=0 and f'(x)=2x+1 in above integral:

now,

Find the arc length for y=x^2+x.

s(x)=\int_{\textcolor{primary}{-1}}^{\textcolor{secondary}{0}}\sqrt{1+(\textcolor{tertiary}{2x+1})^{2}}~dx

Now, we will numerically estimate the integral using the Mid-point rule. Let’s divide the interval [-1,0] into n=2 subintervals.

\Delta x=\frac{0-(-1)}{2}=0.5

Hence, our intervals are:

[-1,-\frac{1}{2}],[-\frac{1}{2},0]

Now, we will find the mid-point of each interval.

m_{1}=\frac{-1+(-\frac{1}{2})}{2}=-0.75

Similarly, using the mid-point formula, the mid-points of the remaining intervals are:

m_{2}=-0.25

Using the Mid-point rule:

M_{2}=\Delta xf(m_{1})+\Delta xf(m_{2})

We have:

f(x)=\sqrt{1+4x^{2}+4x+1}

Now, we will find f(m_{1}). Substitute x=-0.75 in f(x), we get:

f(-0.75)=\sqrt{1+4(-0.75)^{2}+4(-0.75)+1}\approx1.12

Now, we will find f(m_{2}). Substitute x=-0.25 in f(x), we get:

f(-0.25)=\sqrt{1+4(-0.25)^{2}+4(-0.25)+1}\approx1.12

M_{2}=\textcolor{primary}{\Delta x}(f(m_{1})+f(m_{2})

Substitute {\Delta x=\frac{1}{2}} and put the values of f(m_{1}), f(m_{2}).

M_{2}=\textcolor{primary}{\frac{1}{2}}(1.12+1.12)\approx1.12

Hence, the arc length is given by:

s(x)=\int_{-1}^{0}\sqrt{1+4x^{2}+4x+1}dx

For numerical estimation of integral, using the Mid-point rule with 2 subintervals:

S(x)\approx 1.12

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