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Let 
f be the function defined by 
f(x)=4sqrtx. If three subintervals of equal length are used, what is the value of the trapezoidal sum approximation for 
int_(3)^(4.5)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 three subintervals of equal length are used, what is the value of the trapezoidal sum approximation for 34.54xdx \int_{3}^{4.5} 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 three subintervals of equal length are used, what is the value of the trapezoidal sum approximation for 34.54xdx \int_{3}^{4.5} 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 from 33 to 4.54.5 has a length of 4.53=1.54.5 - 3 = 1.5. Since we are using three subintervals, each subinterval will have a width of 1.5/3=0.51.5 / 3 = 0.5.
  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. These points are x=3x = 3, x=3.5x = 3.5, x=4x = 4, and x=4.5x = 4.5. We will calculate f(x)f(x) for each of these xx-values.
  3. Apply Trapezoidal Rule: f(3)=434×1.732=6.928f(3) = 4\sqrt{3} \approx 4 \times 1.732 = 6.928f(3.5)=43.54×1.871=7.484f(3.5) = 4\sqrt{3.5} \approx 4 \times 1.871 = 7.484f(4)=44=4×2=8f(4) = 4\sqrt{4} = 4 \times 2 = 8f(4.5)=44.54×2.121=8.484f(4.5) = 4\sqrt{4.5} \approx 4 \times 2.121 = 8.484
  4. Plug in Values: Now we apply the trapezoidal rule, which is given by the formula:\newlineT=width2(f(a)+2f(a+width)+2f(a+2width)++f(b))T = \frac{\text{width}}{2} \cdot (f(a) + 2\cdot f(a+\text{width}) + 2\cdot f(a+2\cdot \text{width}) + \ldots + f(b))\newlinewhere aa and bb are the start and end points of the interval, and width\text{width} is the width of each subinterval.
  5. Perform Calculations: Plugging in the values we have:\newlineT=(0.52)(f(3)+2f(3.5)+2f(4)+f(4.5))T = (\frac{0.5}{2}) * (f(3) + 2*f(3.5) + 2*f(4) + f(4.5))\newlineT=(0.25)(6.928+2×7.484+2×8+8.484)T = (0.25) * (6.928 + 2\times7.484 + 2\times8 + 8.484)
  6. Round the Result: Now we perform the calculations:\newlineT=(0.25)×(6.928+14.968+16+8.484)T = (0.25) \times (6.928 + 14.968 + 16 + 8.484)\newlineT=(0.25)×(46.38)T = (0.25) \times (46.38)\newlineT11.595T \approx 11.595
  7. Round the Result: Now we perform the calculations:\newlineT=(0.25)×(6.928+14.968+16+8.484)T = (0.25) \times (6.928 + 14.968 + 16 + 8.484)\newlineT=(0.25)×(46.38)T = (0.25) \times (46.38)\newlineT11.595T \approx 11.595We round the result to the nearest thousandth as instructed:\newlineT11.595T \approx 11.595