Approximate the area under the graph of \( f(x)=0.04 x^{4}-1.96 x^{2}+59 \) over the interval \( [3,7] \) by dividing the interval into 4 subintervals. Use the left endpoint of each subinterval. The area under the graph of \( f(x)=0.04 x^{4}-1.96 x^{2}+59 \) over the interval \( [3,7] \) is approximately (Simplify your answer. Type an integer or a decimal.)

To approximate the area under the graph of the function \( f(x) = 0.04 x^{4} - 1.96 x^{2} + 59 \) over the interval \([3, 7]\) using the left endpoint method with 4 subintervals, we first need to determine the width of each subinterval. 1. **Calculate the width of each subinterval:** \[ \text{Width} = \frac{b - a}{n} = \frac{7 - 3}{4} = \frac{4}{4} = 1 \] 2. **Determine the left endpoints of each subinterval:** The subintervals are: - \([3, 4]\) - \([4, 5]\) - \([5, 6]\) - \([6, 7]\) The left endpoints are: - \(x_0 = 3\) - \(x_1 = 4\) - \(x_2 = 5\) - \(x_3 = 6\) 3. **Evaluate the function at each left endpoint:** - \(f(3) = 0.04(3^4) - 1.96(3^2) + 59\) \[ = 0.04(81) - 1.96(9) + 59 = 3.24 - 17.64 + 59 = 44.60 \] - \(f(4) = 0.04(4^4) - 1.96(4^2) + 59\) \[ = 0.04(256) - 1.96(16) + 59 = 10.24 - 31.36 + 59 = 38.88 \] - \(f(5) = 0.04(5^4) - 1.96(5^2) + 59\) \[ = 0.04(625) - 1.96(25) + 59 = 25 - 49 + 59 = 35 \] - \(f(6) = 0.04(6^4) - 1.96(6^2) + 59\) \[ = 0.04(1296) - 1.96(36) + 59 = 51.84 - 70.56 + 59 = 40.28 \] 4. **Calculate the approximate area using the left endpoints:** The approximate area \(A\) is given by: \[ A \approx \text{Width} \times (f(3) + f(4) + f(5) + f(6)) \] Substituting the values we calculated: \[ A \approx 1 \times (44.60 + 38.88 + 35 + 40.28) \] \[ = 44.60 + 38.88 + 35 + 40.28 = 158.76 \] Thus, the approximate area under the graph of \( f(x) \) over the interval \([3, 7]\) is \(\boxed{158.76}\).