In Problems 39-46, approximate each definite integral \( \int_{a}^{b} f(x) d x \) by \( \begin{array}{l}\text { (a) Partitioning }[a, b] \text { into } 4 \text { subintervals and choosing } u_{i} \text { as the left } \\ \text { endpoint of the } i \text { th subinterval. } \\ \text { (b) Partitioning }[a, b] \text { into } 8 \text { subintervals and choosing } u_{i} \text { as the right } \\ \text { endpoint of the ith subinterval. } \\ \text { (4) (c) Use technology to find each integral. } \\ \text { 39. } \int_{0}^{8}\left(x^{2}-4\right) d x\end{array} \)

To approximate the integral \( \int_{0}^{8} (x^2 - 4) \, dx \) using left endpoints with 4 subintervals, we first calculate the width of each subinterval: \( \Delta x = \frac{b-a}{n} = \frac{8-0}{4} = 2 \). The left endpoints will be 0, 2, 4, and 6. Evaluating the function at these points gives us: \[ f(0) = 0^2 - 4 = -4, \quad f(2) = 2^2 - 4 = 0, \quad f(4) = 4^2 - 4 = 12, \quad f(6) = 6^2 - 4 = 32 \] Thus, the approximation is: \[ \int_{0}^{8} (x^2 - 4) \, dx \approx \Delta x \cdot (f(0) + f(2) + f(4) + f(6)) = 2 \cdot (-4 + 0 + 12 + 32) = 2 \cdot 40 = 80 \] For the right endpoint using 8 subintervals, the width is \( \Delta x = \frac{8-0}{8} = 1 \). The right endpoints will be 1, 2, 3, 4, 5, 6, 7, and 8. Evaluating at those points yields: \[ f(1) = 1^2 - 4 = -3, \quad f(2) = 0, \quad f(3) = 5, \quad f(4) = 12, \quad f(5) = 21, \quad f(6) = 32, \quad f(7) = 45, \quad f(8) = 60 \] Thus: \[ \int_{0}^{8} (x^2 - 4) \, dx \approx \Delta x \cdot (f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8)) = 1 \cdot (-3 + 0 + 5 + 12 + 21 + 32 + 45 + 60) \] \[ = 1 \cdot 172 = 172 \] Finally, to find the exact value using technology, we can compute this integral symbolically. The integral computes to: \[ \int_{0}^{8} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{0}^{8} = \left( \frac{8^3}{3} - 4 \cdot 8 \right) - \left( 0 \right) = \left( \frac{512}{3} - 32 \right) = \frac{512}{3} - \frac{96}{3} = \frac{416}{3} \approx 138.67 \]