Let $f(x)=4 x+7, x_{1}=2, x_{2}=4, x_{3}=6, x_{4}=8$, and $\Delta x=2$
(a) Find $\sum_{i=1}^{4} f\left(x_{i}\right) \Delta x$.
(b) The sum in part (a) approximates a definite integral using rectangles. The height of each rectangle is given by the value of the function at the left endpoint. Write the definite integral that the sum approximates.

Step 1 ：Given the function \(f(x)=4x+7\), the values \(x_{1}=2, x_{2}=4, x_{3}=6, x_{4}=8\), and \(\Delta x=2\)

Step 2 ：We need to find \(\sum_{i=1}^{4} f\left(x_{i}\right) \Delta x\)

Step 3 ：This involves substituting the values of \(x_{1}, x_{2}, x_{3}, x_{4}\) into the function \(f(x)\), multiplying each result by \(\Delta x\), and then summing them all up

Step 4 ：Substituting the values into the function gives \(f(x_{1})=4*2+7=15\), \(f(x_{2})=4*4+7=23\), \(f(x_{3})=4*6+7=31\), and \(f(x_{4})=4*8+7=39\)

Step 5 ：Multiplying each result by \(\Delta x\) gives \(15*2=30\), \(23*2=46\), \(31*2=62\), and \(39*2=78\)

Step 6 ：Summing these results gives \(30+46+62+78=216\)

Step 7 ：Thus, \(\sum_{i=1}^{4} f\left(x_{i}\right) \Delta x = 216\)

Step 8 ：\(\boxed{216}\) is the final answer