Part 1
Work through the following steps to approximate $\int_{1}^{10}\left(x^{2}+6\right) d x$.
a) We know that $a=$ and $b=$
10
b) Using 3 subintervals, $\Delta x=3$
Part 20
c) Assume that the sample points in each interval are right endpoints. Find the sample points:
\[
\begin{array}{l}
x_{1}=4 \\
x_{2}=7 \\
x_{3}=10 \quad \square
\end{array}
\]
d) Find the height of each approximating rectangle:
Part 3 of
\[
\begin{array}{l}
f\left(x_{1}\right)=22 \\
f\left(x_{2}\right)=55 \quad \square \\
f\left(x_{3}\right)=106 \quad \square
\end{array}
\]
e) Now find the sum of the areas of 3 approximating rectangles.
Part 4 of 4
\[
\int_{1}^{10} x^{2}+6 d x \approx \sum_{i=1}^{3} f\left(x_{i}\right) \Delta x=
\]

Step 1 ：Given the function \(f(x) = x^2 + 6\), we are asked to approximate the definite integral \(\int_{1}^{10} f(x) dx\) using 3 subintervals and right endpoints.

Step 2 ：First, we calculate the width of each subinterval, which is \(\Delta x = \frac{b - a}{n} = \frac{10 - 1}{3} = 3\).

Step 3 ：Next, we find the sample points, which are the right endpoints of each subinterval. These are \(x_1 = 4\), \(x_2 = 7\), and \(x_3 = 10\).

Step 4 ：Then, we calculate the height of each approximating rectangle, which is the value of the function at each sample point. These are \(f(x_1) = 22\), \(f(x_2) = 55\), and \(f(x_3) = 106\).

Step 5 ：Finally, we find the sum of the areas of the approximating rectangles, which is the approximation of the definite integral. This is \(\int_{1}^{10} f(x) dx \approx \sum_{i=1}^{3} f(x_i) \Delta x = 549\).

Step 6 ：So, the approximation of the definite integral \(\int_{1}^{10}\left(x^{2}+6\right) dx\) using 3 subintervals and right endpoints is \(\boxed{549}\).