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

Ask by Hill Ingram. 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. $$ \int_{0}^{4} \sqrt{x^{2}+2} \mathrm{dx} \approx[?] \quad \mathrm{n}=4 $$

UpStudy AI · Accepted Answer

1. The interval is $$[0,4]$$ and the number of subintervals is $$ n=4 $$. The width of each subinterval is
   $$
   h = \frac{4-0}{4} = 1.
   $$

2. The $$x$$-values at which we will evaluate the function are:
   $$
   x_0 = 0,\quad x_1 = 1,\quad x_2 = 2,\quad x_3 = 3,\quad x_4 = 4.
   $$

3. The function is $$ f(x)=\sqrt{x^2+2} $$. We compute:
   $$
   f(x_0) = \sqrt{0^2+2} = \sqrt{2},
   $$
   $$
   f(x_1) = \sqrt{1^2+2} = \sqrt{3},
   $$
   $$
   f(x_2) = \sqrt{2^2+2} = \sqrt{6},
   $$
   $$
   f(x_3) = \sqrt{3^2+2} = \sqrt{11},
   $$
   $$
   f(x_4) = \sqrt{4^2+2} = \sqrt{18}.
   $$

4. The trapezoidal rule formula is
   $$
   T_n = \frac{h}{2} \left[f(x_0)+f(x_4)+2\left(f(x_1)+f(x_2)+f(x_3)\right)\right].
   $$
   Substituting in the values, we get:
   $$
   T_4 = \frac{1}{2} \left[\sqrt{2}+\sqrt{18}+2\left(\sqrt{3}+\sqrt{6}+\sqrt{11}\right)\right].
   $$

5. Approximating the square roots:
   $$
   \sqrt{2} \approx 1.414,\quad \sqrt{3} \approx 1.732,\quad \sqrt{6} \approx 2.449,\quad \sqrt{11} \approx 3.317,\quad \sqrt{18} \approx 4.243.
   $$

6. Substitute these approximations:
   $$
   T_4 \approx \frac{1}{2} \left[1.414 + 4.243 + 2(1.732 + 2.449 + 3.317)\right].
   $$
   Compute the sum inside the parentheses of the doubled terms:
   $$
   1.732 + 2.449 + 3.317 = 7.498.
   $$
   Then,
   $$
   2(7.498) = 14.996.
   $$
   Now, add the endpoints:
   $$
   1.414 + 4.243 = 5.657.
   $$
   So,
   $$
   T_4 \approx \frac{1}{2} \left[5.657 + 14.996\right] = \frac{1}{2} \times 20.653 = 10.327.
   $$

7. Therefore, the trapezoidal rule approximation of the integral is
   $$
   \boxed{10.327}.
   $$
The trapezoidal rule approximation of the integral $$ \int_{0}^{4} \sqrt{x^{2}+2} \, dx $$ with $$ n=4 $$ trapezoids is approximately **10.327**.

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.