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Let 
f be the function defined by 
f(x)=4sqrtx. If four subintervals of equal length are used, what is the value of the trapezoidal sum approximation for 
int_(1)^(7)4sqrtxdx ? Round to the nearest thousandth if necessary.
Answer:

Let f f be the function defined by f(x)=4x f(x)=4 \sqrt{x} . If four subintervals of equal length are used, what is the value of the trapezoidal sum approximation for 174xdx \int_{1}^{7} 4 \sqrt{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:

Full solution

Q. Let f f be the function defined by f(x)=4x f(x)=4 \sqrt{x} . If four subintervals of equal length are used, what is the value of the trapezoidal sum approximation for 174xdx \int_{1}^{7} 4 \sqrt{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Determine Subinterval Width: To use the trapezoidal rule, we first need to determine the width of each subinterval. The interval [1,7][1, 7] has a length of 71=67 - 1 = 6. Since we are using four subintervals, each subinterval will have a width of 6/4=1.56 / 4 = 1.5.
  2. Calculate Function Values: Next, we need to calculate the values of the function f(x)=4xf(x) = 4\sqrt{x} at the endpoints of each subinterval. The endpoints are x=1x = 1, x=2.5x = 2.5, x=4x = 4, x=5.5x = 5.5, and x=7x = 7. We will calculate f(x)f(x) for each of these xx-values.
  3. Apply Trapezoidal Rule: f(1)=41=4(1)=4f(1) = 4\sqrt{1} = 4(1) = 4f(2.5)=42.54(1.581)6.324f(2.5) = 4\sqrt{2.5} \approx 4(1.581) \approx 6.324f(4)=44=4(2)=8f(4) = 4\sqrt{4} = 4(2) = 8f(5.5)=45.54(2.345)9.38f(5.5) = 4\sqrt{5.5} \approx 4(2.345) \approx 9.38f(7)=474(2.646)10.584f(7) = 4\sqrt{7} \approx 4(2.646) \approx 10.584
  4. Plug in Values: Now we apply the trapezoidal rule, which is given by the formula:\newlineT=width2(f(x0)+2f(x1)+2f(x2)++2f(xn1)+f(xn))T = \frac{\text{width}}{2} \cdot (f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n))\newlinewhere x0,x1,,xnx_0, x_1, \ldots, x_n are the endpoints of the subintervals and nn is the number of subintervals.
  5. Perform Calculations: Plugging in the values we have:\newlineT=1.52×(f(1)+2f(2.5)+2f(4)+2f(5.5)+f(7))T = \frac{1.5}{2} \times (f(1) + 2f(2.5) + 2f(4) + 2f(5.5) + f(7))\newlineT=(0.75)×(4+2(6.324)+2(8)+2(9.38)+10.584)T = (0.75) \times (4 + 2(6.324) + 2(8) + 2(9.38) + 10.584)
  6. Round to Nearest Thousandth: Now we perform the calculations:\newlineT=0.75×(4+12.648+16+18.76+10.584)T = 0.75 \times (4 + 12.648 + 16 + 18.76 + 10.584)\newlineT=0.75×(61.992)T = 0.75 \times (61.992)\newlineT46.494T \approx 46.494
  7. Round to Nearest Thousandth: Now we perform the calculations:\newlineT=0.75×(4+12.648+16+18.76+10.584)T = 0.75 \times (4 + 12.648 + 16 + 18.76 + 10.584)\newlineT=0.75×(61.992)T = 0.75 \times (61.992)\newlineT46.494T \approx 46.494Rounding to the nearest thousandth, we get:\newlineT46.494T \approx 46.494