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 \(\boxed{549}\).