Approximate the value of the integral ∫ (cos(x))/x dx (with lower bound, 1 and upper bound, 5) A.) by using the Trapezoidal Rule T8 B.) using the Midpoint Rule M8 C.) by using the Simpson’s rule S8

Answer 1

With a lower limit of 1 and an upper limit of 5, since 5 - 1 = 4 and 4/8 = 0.5, the width of each rectangle is 0.5.

For the trapezoidal rule, calculate the function value at each side of the rectangle (so for the first rectangle, do f(1) and f(1.5)) and average them together. That is the rectangle height. f(1) = 0.5403 and f(1.5) = 0.0472. Averaging them gives 0.2937. Multiply by the rectangle width to get the area. (0.5)(0.2937) = 0.1469. Then repeat for each of the other 7 rectangles.

For the midpoint rule, take f(midpoint) which for the first triangle would be f(1.25) which is 0.2523 and multiply that by the width (0.5) so the area of the 1st rectangle is 0.1261. And repeat for the other 7 rectangles.

Simpson's Rule's rule say _a∫^bf(x)dx ≈Δx/3[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + 2f(x₄) + . . . +f(x_n)] where the middle alternates between the coefficients 4 and 2.

In our case Δx = 0.5 so the integral ≈

0.5/3[f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + 2f(3) + 4f(3.5) + 2f(4) + 4f(4.5) + f(5)]

1/6[0.5403 + 4(0.0472) + 2(-0.2081) + 4(-0.3205) + 2(-0.3300) + 4(-0.2676) + 2(-0.1634)+ 4(-0.0468) + 0.0567] = 1/6(-3.1567) = -0.5261