Question

Use the Trapezoidal Rule to approximate ∫73x2+6‾‾‾‾‾‾√dx using n=3. Round your answer to the nearest hundredth.

Answer 1

Answer: 22.39

Given:

$\int_3^7\sqrt{x^2+6}dx\text{ }n=3$

The trapezoidal rule states that:

$\int_a^bf(x)dx\approx\frac{\Delta x}{2}(f(x_0)+2f(x_1)+2f(x_2)+2f(x3)+...+f(x_n)$

Where:

$\Delta x=\frac{b-a}{n}$

From the given, we know that:

$\begin{gathered} f(x)=\sqrt{x^2+6} \\ a=3 \\ b=7 \\ n=3 \end{gathered}$

With this, we know that:

$\Delta x=\frac{7-3}{3}=\frac{4}{3}$

We will then divide the interval [3,7] into n=3 subintervals of 4/3, which will give us:

$3,\frac{13}{3},\frac{17}{3},7$

Now, we will evaluate the function at these endpoints:

$\begin{gathered} f(x)=\sqrt{x^2+6} \\ \Rightarrow f(3)=\sqrt{(3)^2+6}=\sqrt{15} \\ \operatorname{\Rightarrow}2f(\frac{13}{3})=2\sqrt{(\frac{13}{3})^2+6}=\frac{2\sqrt{223}}{3} \\ \operatorname{\Rightarrow}2f(\frac{17}{3})=2\sqrt{(\frac{17}{3})^2+6}=\frac{14\sqrt{7}}{3} \\ \Rightarrow f(7)=\sqrt{(7)^2+6}=\sqrt{55}\frac{}{} \end{gathered}$

We will then sum up the values and multiply by Δx/2:

$\begin{gathered} \frac{\Delta x}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3} \\ \Rightarrow\frac{2}{3}(\sqrt{15}+\frac{2\sqrt{223}}{3}+\frac{14\sqrt{7}}{3}+\sqrt{55}) \\ =22.3943\approx22.39 \end{gathered}$

Therefore,

$\int_3^7\sqrt{x^2+6}dx\approx22.39$