which of the following is the midpoint riemann sum approximation of ∫₄⁶ √(x³ + 1)dx using 4 subintervals of…
which of the following is the midpoint riemann sum approximation of ∫₄⁶ √(x³ + 1)dx using 4 subintervals of equal width? (a) ¼(√(4.25³ + 1)+√(4.75³ + 1)+√(5.25³ + 1)+√(5.75³ + 1)) (b) ½(√(4.25³ + 1)+√(4.75³ + 1)+√(5.25³ + 1)+√(5.75³ + 1)) (c) ¼((√(4³+1)+√(4.5³+1))/2 + (√(4.5³+1)+√(5³+1))/2 + (√(5³+1)+√(5.5³+1))/2 + (√(5.5³+1)+√(6³+1))/2) (d) ½((√(4³+1)+√(4.5³+1))/2 + (√(4.5³+1)+√(5³+1))/2 + (√(5³+1)+√(5.5³+1))/2 + (√(5.5³+1)+√(6³+1))/2)
Answer
Explanation:
Step1: Calculate width of sub - intervals
The interval is from $a = 4$ to $b=6$ and $n = 4$. The width $\Delta x=\frac{b - a}{n}=\frac{6 - 4}{4}=\frac{1}{2}$.
Step2: Determine mid - points of sub - intervals
The sub - intervals are $[4,4.5],[4.5,5],[5,5.5],[5.5,6]$. The mid - points are $x_1 = 4.25,x_2=4.75,x_3 = 5.25,x_4=5.75$.
Step3: Write mid - point Riemann sum formula
The mid - point Riemann sum $M_n=\Delta x\sum_{i = 1}^{n}f(x_i^)$, where $x_i^$ are mid - points and $\Delta x$ is the width of sub - intervals. Here, $f(x)=\sqrt{x^{3}+1}$, $\Delta x=\frac{1}{2}$, and $n = 4$. So $M_4=\frac{1}{2}(\sqrt{4.25^{3}+1}+\sqrt{4.75^{3}+1}+\sqrt{5.25^{3}+1}+\sqrt{5.75^{3}+1})$.
Answer:
B. $\frac{1}{2}(\sqrt{4.25^{3}+1}+\sqrt{4.75^{3}+1}+\sqrt{5.25^{3}+1}+\sqrt{5.75^{3}+1})$