### Problem Statement**3. Estimate the integral \( \int_{-2}^{6} (x^3 - 18x^2 + 107x - 207) \, dx \) using a left-hand Riemann sum with 4 subintervals of equal width.**### ExplanationTo solve this problem, we will use the left-hand Riemann sum to approximate the definite integral of the given function over the interval \([-2, 6]\). The interval will be divided into 4 equal subintervals, and the left endpoint of each subinterval will be used to evaluate the function.1. **Determine the Width of Each Subinterval**: - The total interval length is \(6 - (-2) = 8\). - Since there are 4 subintervals, the width \(\Delta x\) of each subinterval is \( \frac{8}{4} = 2\).2. **Identify the Left Endpoints of Each Subinterval**: - Subintervals and their left endpoints: - \([-2, 0]\), left endpoint: \(-2\) - \([0, 2]\), left endpoint: \(0\) - \([2, 4]\), left endpoint: \(2\) - \([4, 6]\), left endpoint: \(4\)3. **Evaluate the Function at Each Left Endpoint**: - \(f(-2) = (-2)^3 - 18(-2)^2 + 107(-2) - 207\) - \(f(0) = (0)^3 - 18(0)^2 + 107(0) - 207\) - \(f(2) = (2)^3 - 18(2)^2 + 107(2) - 207\) - \(f(4) = (4)^3 - 18(4)^2 + 107(4) - 207\)4. **Calculate the Left-Hand Riemann Sum**: - Riemann sum = \(\Delta x \times [f(-2) + f(0) + f(2) + f(4)]\) - Substitute the values to approximate the integral.This approach provides an approximation of the area under the curve of the function from \(-2\) to

Consider the following expression: ∫-26x3-18x2+107x-207dx n=4 The interval is -2 to 6. The subintervals are -2,0,2,4,6. Hence, the left endpoints are ξi=0,2,4,6.The values of the function fξi at the left endpoints: fξ1=f0=03-18×02+107×0-207=-207 fξ2=f2=23-18×22+107×2-207=-57 fξ3=f4=43-18×42+107×4-207=-3 fξ4=f6=63-18×62+107×6-207=3The left Riemann sum with n=4: L4=∑i=14fξi∆x where ∆x=2. L4=2×-207-57-3+3=2×-264=-528 Hence, the solution is ∫-26x3-18x2+107x-207dx≈-528.