Compute 0^1 (1+x^2)^ 1 dx , by Simpson's 13 rule. Also compare it with true solution. [6] OR Approximate the value for 1^1 e^x dx using Trapezoidal rule with n = 2 .

Solution 1: Compute ${\int }_0^1(1+x^2{)}^{-1}dx$ by Simpson’s $\frac{1}{3}$ Rule and Compare with True Solution

Step 1: Understand Simpson’s $\frac{1}{3}$ Rule

Simpson’s $\frac{1}{3}$ rule approximates the integral of a function $f(x)$ over $[a,b]$ by fitting a quadratic polynomial to the function at three points: $x_0=a$, $x_1=\frac{a+b}{2}$, and $x_2=b$. The formula is:

$${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[f(x_0)+4f(x_1)+f(x_2)\right],$$

where $h=\frac{b-a}{2}$.

Step 2: Apply Simpson’s Rule to ${\int }_0^1\frac{1}{1+x^2}dx$

Given:

• $a=0$, $b=1$
• $f(x)=\frac{1}{1+x^2}$
• $n=2$ (since Simpson’s rule requires an even number of subintervals, but here we use the basic form with 2 subintervals, i.e., 3 points).

Compute $h$: $$h=\frac{b-a}{2}=\frac{1-0}{2}=0.5$$

Compute the points: $$x_0=0,x_1=0.5,x_2=1$$

Compute $f(x)$ at these points: $$f(x_0)=f(0)=\frac{1}{1+0^2}=1$$ $$f(x_1)=f(0.5)=\frac{1}{1+(0.5{)}^2}=\frac{1}{1.25}=0.8$$ $$f(x_2)=f(1)=\frac{1}{1+1^2}=\frac{1}{2}=0.5$$

Apply Simpson’s rule: $${\int }_0^1\frac{1}{1+x^2}dx\approx \frac{0.5}{3}\left[1+4(0.8)+0.5\right]$$ $$=\frac{0.5}{3}\left[1+3.2+0.5\right]$$ $$=\frac{0.5}{3}\times 4.7$$ $$=\frac{2.35}{3}\approx 0.7833$$

Step 3: Compare with the True Solution

The true value of the integral is: $${\int }_0^1\frac{1}{1+x^2}dx=\arctan (x){|}_0^1=\arctan (1)-\arctan (0)=\frac{\pi }{4}\approx 0.7854$$

Step 4: Comparison Table

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Solution 2: Approximate ${\int }_{-1}^1{e}^{x}dx$ Using Trapezoidal Rule with $n=2$

Step 1: Understand the Trapezoidal Rule

The trapezoidal rule approximates the integral of $f(x)$ over $[a,b]$ by dividing the interval into $n$ subintervals and summing the areas of trapezoids formed under the curve. The formula is:

$${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots +2f(x_{n-1})+f(x_n)\right],$$

where $h=\frac{b-a}{n}$.

Step 2: Apply the Trapezoidal Rule to ${\int }_{-1}^1{e}^{x}dx$ with $n=2$

Given:

• $a=-1$, $b=1$
• $f(x)=e^x$
• $n=2$

Compute $h$: $$h=\frac{b-a}{n}=\frac{1-(-1)}{2}=1$$

Compute the points: $$x_0=-1,x_1=0,x_2=1$$

Compute $f(x)$ at these points: $$f(x_0)=f(-1)=e^{-1}\approx 0.3679$$ $$f(x_1)=f(0)=e^0=1$$ $$f(x_2)=f(1)=e^1\approx 2.7183$$

Apply the trapezoidal rule: $${\int }_{-1}^1{e}^{x}dx\approx \frac{1}{2}\left[f(-1)+2f(0)+f(1)\right]$$ $$=\frac{1}{2}\left[0.3679+2(1)+2.7183\right]$$ $$=\frac{1}{2}\left[0.3679+2+2.7183\right]$$ $$=\frac{1}{2}\times 5.0862\approx 2.5431$$

Step 3: Compare with the True Solution

The true value of the integral is: $${\int }_{-1}^1{e}^{x}dx=e^x{|}_{-1}^1=e^1-e^{-1}\approx 2.7183-0.3679=2.3504$$

Step 4: Comparison Table

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Final Answers

1. Simpson’s $\frac{1}{3}$ Rule for ${\int }_0^1(1+x^2{)}^{-1}dx$: $$\boxed{0.7833}$$

2. Trapezoidal Rule for ${\int }_{-1}^1{e}^{x}dx$ with $n=2$: $$\boxed{2.5431}$$