Recall the Trapezoidal Rule where \Delta x = b − a n and xi = a + i\Delta x. b a f(x) dx ≈ Tn = \Delta x 2 f(x0) + 2f(x1) + + 2f(xn − 1) + f(xn) We need to use the Trapezoidal Rule to estimate the following with n = 4 subintervals. 1 0 4 cos(x2) dx We first determine the following values. a = b = \Delta x =
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The Trapezoidal Rule says that ∫[a to b] f(x) dx ≈ Tn = Δx/2 [f(x0) + 2f(x1) + ... + 2f(xn-1) + f(xn)]. We need to estimate ∫[0 to 1] 10 cos(x²) dx with n = 4 subintervals. We have Δx = 0.25 Therefore, Δx/2 = 0.125 Step 2 We know that x0 represents the beginning of the first subinterval, so x0 = 0, and x1 is the endpoint of the first sub-interval (which is also the beginning of the second subinterval), and so x1 = 0.25. Step 3 This gives us T4 = 1/8 [10 cos(0) + 20 cos(1/16) + 20 cos(1/4) + 20 cos(9/16) + 10 cos(1)] = (rounded to six decimal places) Therefore, using the Trapezoidal Rule with n = 4 and rounding to six decimal places we have ∫[0 to 1] 10 cos(x²) dx ≈
Krishna S.
The Trapezoidal Rule says that ∫(b to a) f(x) dx ≈ Tn = (Δx/2) [f(x0) + 2f(x1) + ... + 2f(xn-1) + f(xn)]. We again need to estimate ∫(0 to 1/6) cos(x^2) dx with n = 4 subintervals. We have Δx = 1/8. Therefore, T4 = (1/8) [f(0) + 2f(1/4) + 2f(1/2) + 2f(3/4) + f(1)].
Adi S.
the Midpoint Rule The Midpoint Rule says that ∫[a,b] f(x) dx ≈ Mn = Δx[f(x̄₁) + f(x̄₂) + ... + f(x̄n)] with Δx = (b - a) / n. We again need to estimate ∫[0,1] 4 cos(x²) dx with n = 4 subintervals. For this, Δx = (1 - 0) / 4 = 1/4. We know that x̄₁ represents the midpoint of the first subinterval [0, 1/4]. The midpoint of [0, 1/4] is x̄₁ = (1/4 + 0) / 2 = 0.125. Our function is f(x) = 4 cos(x²). So we have f(x̄₁) = 4 cos( ). Submit Skip (you cannot come back)
Sri K.
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