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Let 
f be the function defined by 
f(x)=2^(x). If three subintervals of equal length are used, what is the value of the right Riemann sum approximation for 
int_(1)^(2.5)2^(x)dx ? Round to the nearest thousandth if necessary.
Answer:

Let f f be the function defined by f(x)=2x f(x)=2^{x} . If three subintervals of equal length are used, what is the value of the right Riemann sum approximation for 12.52xdx \int_{1}^{2.5} 2^{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:

Full solution

Q. Let f f be the function defined by f(x)=2x f(x)=2^{x} . If three subintervals of equal length are used, what is the value of the right Riemann sum approximation for 12.52xdx \int_{1}^{2.5} 2^{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Determine Subinterval Length: We are given the function f(x)=2xf(x) = 2^x and we need to approximate the integral from x=1x = 1 to x=2.5x = 2.5 using a right Riemann sum with three subintervals of equal length. First, we need to determine the length of each subinterval.\newlineThe total interval length is 2.51=1.52.5 - 1 = 1.5. Since we are using three subintervals, the length of each subinterval is 1.53=0.5\frac{1.5}{3} = 0.5.
  2. Find Right Endpoints: Next, we need to find the right endpoints of each subinterval. Since we start at x=1x = 1 and each subinterval is 0.50.5 units long, the right endpoints will be at x=1.5x = 1.5, x=2x = 2, and x=2.5x = 2.5.
  3. Evaluate Function at Endpoints: Now we evaluate the function f(x)=2xf(x) = 2^x at each of the right endpoints. This gives us the heights of the rectangles for the Riemann sum.\newlinef(1.5)=21.5f(1.5) = 2^{1.5}\newlinef(2)=22f(2) = 2^{2}\newlinef(2.5)=22.5f(2.5) = 2^{2.5}
  4. Calculate Function Values: We calculate the values of the function at these points:\newlinef(1.5)=21.52.8284f(1.5) = 2^{1.5} \approx 2.8284\newlinef(2)=22=4f(2) = 2^{2} = 4\newlinef(2.5)=22.55.6569f(2.5) = 2^{2.5} \approx 5.6569
  5. Calculate Riemann Sum: The right Riemann sum is the sum of the areas of the rectangles. Each rectangle has a base of 0.50.5 (the subinterval length) and a height given by the function value at the right endpoint. So the Riemann sum is:\newlineRiemann sum 0.5×f(1.5)+0.5×f(2)+0.5×f(2.5)\approx 0.5 \times f(1.5) + 0.5 \times f(2) + 0.5 \times f(2.5)\newlineRiemann sum 0.5×2.8284+0.5×4+0.5×5.6569\approx 0.5 \times 2.8284 + 0.5 \times 4 + 0.5 \times 5.6569
  6. Perform Calculations: Performing the calculations gives us:\newlineRiemann sum 0.5×2.8284+0.5×4+0.5×5.6569\approx 0.5 \times 2.8284 + 0.5 \times 4 + 0.5 \times 5.6569\newlineRiemann sum 1.4142+2+2.8284\approx 1.4142 + 2 + 2.8284\newlineRiemann sum 6.2426\approx 6.2426
  7. Round to Nearest Thousandth: Rounding to the nearest thousandth, we get: Riemann sum6.243\text{Riemann sum} \approx 6.243

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