
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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