Question

1. using the following matlab script as a guide, fix the script to evaluate the function 4x + 6x^3 from -2 to 2 using 300 steps. a = 0; b = 0; n = 0; h = (b - a)/n; integral = 0; for x = a:h:b - h y1 = x; y2 = x; integral = integral + 0.5h(y1 + y2); end integral (long answer) *

Explanation:

Step1: Set the correct values for a, b, and n

The function is evaluated from -2 to 2, so \(a=-2\), \(b = 2\). The number of steps \(n = 300\).

Step2: Calculate the step size h

The formula for the step - size is \(h=\frac{b - a}{n}\). Substituting \(a=-2\), \(b = 2\), and \(n = 300\), we get \(h=\frac{2-(-2)}{300}=\frac{4}{300}=\frac{1}{75}\).

Step3: Define the function values

The function is \(y = 4x+6x^{3}\). In the loop, for each \(x\) value (from \(a\) to \(b - h\) with step \(h\)), \(y1=4x + 6x^{3}\) and \(y2=4(x + h)+6(x + h)^{3}\).

Step4: Implement the trapezoidal rule formula

The trapezoidal rule formula for approximating the integral \(\int_{a}^{b}f(x)dx\approx\sum_{i = 0}^{n-1}\frac{h}{2}(f(x_{i})+f(x_{i + 1}))\). So, in the loop, the integral is updated as \(integral=integral + 0.5h(y1 + y2)\).

The corrected MATLAB script is:

a=-2;
b = 2;
n = 300;
h=(b - a)/n;
integral = 0;
for x=a:h:b - h
    y1=4*x+6*x^3;
    y2=4*(x + h)+6*(x + h)^3;
    integral=integral + 0.5*h*(y1 + y2);
end
integral

Answer:

The corrected MATLAB script is:

a=-2;
b = 2;
n = 300;
h=(b - a)/n;
integral = 0;
for x=a:h:b - h
    y1=4*x+6*x^3;
    y2=4*(x + h)+6*(x + h)^3;
    integral=integral + 0.5*h*(y1 + y2);
end
integral