Use the trapezoidal rule to
approximate the integral with
trapezoids. Round your answer
to the nearest thousandth.

Ask by Floyd Griffin. in the United States

Mar 22,2025

Question

Use the trapezoidal rule to approximate the integral with $$ n $$ trapezoids. Round your answer to the nearest thousandth. $$ \int_{2}^{8} \sqrt{x^{3}+4} \mathrm{dx} \approx[?] \quad \mathrm{n}=3 $$

UpStudy AI · Accepted Answer

Let $$ f(x)=\sqrt{x^3+4} $$.

**Step 1. Compute step width $$ h $$:**

$$
h=\frac{8-2}{3}=\frac{6}{3}=2
$$

**Step 2. Identify the x-values:**

$$
x_0 = 2,\quad x_1 = 2+2=4,\quad x_2 = 4+2=6,\quad x_3 = 8
$$

**Step 3. Evaluate the function at these points:**

$$
f(2)=\sqrt{2^3+4}=\sqrt{8+4}=\sqrt{12} \approx 3.464
$$
$$
f(4)=\sqrt{4^3+4}=\sqrt{64+4}=\sqrt{68} \approx 8.246
$$
$$
f(6)=\sqrt{6^3+4}=\sqrt{216+4}=\sqrt{220} \approx 14.832
$$
$$
f(8)=\sqrt{8^3+4}=\sqrt{512+4}=\sqrt{516} \approx 22.716
$$

**Step 4. Apply the trapezoidal rule:**

The formula for $$ n $$ subintervals is

$$
T_n=\frac{h}{2} \Big[ f(x_0) + f(x_n) \Big] + h \sum_{i=1}^{n-1} f(x_i)
$$

For $$ n=3 $$:

$$
T_3=\frac{2}{2}\Big[f(2)+f(8)\Big]+2\Big[f(4)+f(6)\Big]
$$
$$
T_3=1\Big[\sqrt{12}+\sqrt{516}\Big]+2\Big[\sqrt{68}+\sqrt{220}\Big]
$$
$$
T_3\approx 1\Big[3.464+22.716\Big]+2\Big[8.246+14.832\Big]
$$
$$
T_3\approx 26.180+2(23.078)
$$
$$
T_3\approx 26.180+46.156=72.336
$$

**Final Answer:**

$$
\boxed{72.336}
$$
The approximate value of the integral is 72.336.

Use the trapezoidal rule to
approximate the integral with
trapezoids. Round your answer
to the nearest thousandth.

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Use the trapezoidal rule to approximate the integral with trapezoids. Round your answer to the nearest thousandth.