Approximate the area under the graph of \( f(x)=0.03 x^{4}-1.21 x^{2}+64 \) over the interval \( [2,10] \) by dividing the interval into 4 subintervals. Use the left endpoint of each subinterval. The area under the graph of \( f(x)=0.03 x^{4}-1.21 x^{2}+64 \) over the interval \( [2,10] \) is approximately (Simplify your answer. Type an integer or a decimal.)

To approximate the area under the graph of the function \( f(x)=0.03x^{4}-1.21x^{2}+64 \) over the interval \([2,10]\) by dividing the interval into 4 subintervals using the left endpoint of each subinterval, we can use the Riemann sum formula. The Riemann sum formula for approximating the area under a curve is given by: \[ \text{Area} \approx \Delta x \times \left[ f(x_1) + f(x_2) + f(x_3) + f(x_4) \right] \] where: - \( \Delta x \) is the width of each subinterval, - \( x_1, x_2, x_3, x_4 \) are the left endpoints of the subintervals. Given that the interval is \([2,10]\) and we are dividing it into 4 subintervals, the width of each subinterval is: \[ \Delta x = \frac{10-2}{4} = 2 \] The left endpoints of the subintervals are: - \( x_1 = 2 \) - \( x_2 = 4 \) - \( x_3 = 6 \) - \( x_4 = 8 \) Now, we can calculate the Riemann sum by substituting the values into the formula: \[ \text{Area} \approx 2 \times \left[ f(2) + f(4) + f(6) + f(8) \right] \] Let's calculate the values of \( f(2) \), \( f(4) \), \( f(6) \), and \( f(8) \) first. Calculate the value by following steps: - step0: Calculate: \(0.03\times 2^{4}-1.21\times 2^{2}+64\) - step1: Multiply the numbers: \(0.48-1.21\times 2^{2}+64\) - step2: Multiply the numbers: \(0.48-4.84+64\) - step3: Calculate: \(59.64\) Calculate or simplify the expression \( 0.03*4^4-1.21*4^2+64 \). Calculate the value by following steps: - step0: Calculate: \(0.03\times 4^{4}-1.21\times 4^{2}+64\) - step1: Multiply the numbers: \(7.68-1.21\times 4^{2}+64\) - step2: Multiply the numbers: \(7.68-19.36+64\) - step3: Calculate: \(52.32\) Calculate or simplify the expression \( 0.03*6^4-1.21*6^2+64 \). Calculate the value by following steps: - step0: Calculate: \(0.03\times 6^{4}-1.21\times 6^{2}+64\) - step1: Multiply the numbers: \(38.88-1.21\times 6^{2}+64\) - step2: Multiply the numbers: \(38.88-43.56+64\) - step3: Calculate: \(59.32\) Calculate or simplify the expression \( 0.03*8^4-1.21*8^2+64 \). Calculate the value by following steps: - step0: Calculate: \(0.03\times 8^{4}-1.21\times 8^{2}+64\) - step1: Multiply the numbers: \(122.88-1.21\times 8^{2}+64\) - step2: Multiply the numbers: \(122.88-77.44+64\) - step3: Calculate: \(109.44\) Now, we have calculated the values of \( f(2) \), \( f(4) \), \( f(6) \), and \( f(8) \): - \( f(2) = 59.64 \) - \( f(4) = 52.32 \) - \( f(6) = 59.32 \) - \( f(8) = 109.44 \) Next, we can substitute these values into the Riemann sum formula to approximate the area under the graph of the function over the interval \([2,10]\). Calculate the value by following steps: - step0: Calculate: \(2\left(59.64+52.32+59.32+109.44\right)\) - step1: Add the numbers: \(2\times 280.72\) - step2: Multiply the numbers: \(561.44\) The approximate area under the graph of the function \( f(x)=0.03x^{4}-1.21x^{2}+64 \) over the interval \([2,10]\) by dividing the interval into 4 subintervals using the left endpoint of each subinterval is approximately 561.44 square units.