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"Estimate the integral ∫301x^3 + 1−−−−−√dx by the trapezoidal rule using n = 4."
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Step 1:
: Understand the problem and gather necessary information

We are asked to estimate the integral ∫(301)x^3 + 1 dx from 0 to 1 using the trapezoidal rule with n = 4.

Step 2:
: Set up the trapezoidal rule formula

∫(f(x)) dx ≈ T_n = rac{Δx}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]
The trapezoidal rule formula is given by: where Δx = (b - a) / n, and in our case, a = 0, b = 1, and n = 4.

Step 3:
: Calculate Δx

Δx = rac{(b - a)}{n} = rac{1 - 0}{4} = rac{1}{4}

Step 4:
: Calculate the function values at the given points

f(x_4) = (301)(1)3+1 = 302
We need to find the function values at x^0 = 0, x^1 = Δx, x^2 = 2Δx, x^3 = 3Δx, and x^4 = 1. So,

Step 5:
: Apply the trapezoidal rule formula

T_4 = rac{1/4}{2} [1 + 2(1.4645) + 2(2.8780) + 2(4.2400) + 302]
Now, we can apply the trapezoidal rule formula: Substitute the values:

Step 6:
: Calculate the approximate integral value

T_4 ≈ rac{1/4}{2} (1 + 2(1.4645) + 2(2.8780) + 2(4.2400) + 302) ≈ 126.1573
Calculate the sum and multiply by Δx/ 2:

Final Answer

The approximate value of the integral ∫(301)x^3 + 1 dx from 0 to 1 using the trapezoidal rule with n = 4 is approximately 126.1573.