Given: `dx 3. Evaluate √1+x² 2 using Trapezoidal rule with h = 0.2. 0`The given equation is `√1 + x²`Interval `a = 0` and `b = 2`.Trapezoidal rule: `∫ a b f(x) dx = h/2 [f(x₀) + 2(f(x₁) + .....+ f(x(n-1))) + f(xn)]`where `h = (b-a)/n` and `x₀ = a, x₁ = a + h, x₂ = a + 2h, ......, xn = b`Trapezoidal Rule for this equation is: `∫₀² √1 + x² dx ≈ h/2 [f(0) + 2(f(0.2) + f(0.4) + f(0.6) + f(0.8) + f(1.0) + f(1.2) + f(1.4) + f(1.6) + f(1.8) + f(2.0))]`Where `h = 0.2`=`0.2/2`[ `f(0)`+`2(f(0.2) + f(0.4) + f(0.6) + f(0.8) + f(1.0) + f(1.2) + f(1.4) + f(1.6) + f(1.8)` + `f(2)` ]`= 0.1[ f(0) + 2(f(0.2) + f(0.4) + f(0.6) + f(0.8) + f(1.0) + f(1.2) + f(1.4) + f(1.6) + f(1.8) + f(2) ]`We have to find the value of `f(x)` as `√1 + x²` at each `x` point.Substituting the values in the equation, we get `f(x)`: `f(0) = √1 + 0² = 1` `f(0.2) = √1 + 0.2² = 1.00499` `f(0.4) = √1 + 0.4² = 1.0198` `f(0.6) = √1 + 0.6² = 1.04212` `f(0.8) = √1 + 0.8² = 1.07414` `f(1.0) = √1 + 1² = 1.11803` `f(1.2) = √1 + 1.2² = 1.17639` `f(1.4) = √1 + 1.4² = 1.25283` `f(1.6) = √1 + 1.6² = 1.35164` `f(1.8) = √1 + 1.8² = 1.47925` `f(2) = √1 + 2² = 2.236`Plugging all the values in the above formula we get:`0.1[1 + 2(1.00499 + 1.0198 + 1.04212 + 1.07414 + 1.11803 + 1.17639 + 1.25283 + 1.35164 + 1.47925) + 2.236]`=`0.1 [1 + 20.1094 + 2.236]`=`0.1 (23.3454)`=`2.33454`Therefore, the main answer is `2.33454`As the second question is separate, let's answer it:2. Solve the system of equations `x - 2y = 0` and `2x + y = 5` by `4(2)`Adding these equations, we get: `(x - 2y) + (2x + y) = 0 + 5`On solving we get: `3x - y = 5`Multiplying the second equation by 2, we get: `2(2x + y) = 2(5)`On solving we get: `4x + 2y = 10`Divide the equation by 2 we get: `2x + y = 5`This equation is same as we got while adding the two given equations.We have solved the system of equations using substitution method. The solution is `x = 5/3` and `y = 5/3`.Hence, the conclusion is `Trapezoidal Rule for given equation is 2.33454 and the solution of the given system of equations is x = 5/3 and y = 5/3.`