Suppose S₆is the upper sum of the area enclosed by the graph $f ( x ) = 10 - x ^ { 2 },$ the x-axis, and the lines x = 0 and x = 3 by partitioning [0, 3] into six subintervals [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3]. Then S₆ is

Question

Suppose S₆is the upper sum of the area enclosed by the graph

f ( x ) = 10 - x ^ { 2 },

the x-axis, and the lines x = 0 and x = 3 by partitioning [0, 3] into six subintervals [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3]. Then S₆ is

Quizplus · Accepted Answer

The upper sum for a function $f(x) = 10 - x^2$ over the interval [0, 3] with subintervals [0, 0.5], [0.5, 1], [1, 1.5], [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3] is calculated by taking the sum of the areas of rectangles, where the height of each rectangle is the maximum value of the function in the subinterval. The width of each subinterval is 0.5. For each subinterval, calculate the maximum value of $f(x)$ (which will be at the left endpoint for this downward-opening parabola), and then multiply by the width (0.5) to find the area of each rectangle. Sum these areas to find the total upper sum.1. [0, 0.5]: Max value at x=0, $f(0) = 10$. Area = $10 \times 0.5 = 5$2. [0.5, 1]: Max value at x=0.5, $f(0.5) = 10 - (0.5)^2 = 9.75$. Area = $9.75 \times 0.5 = 4.875$3. [1, 1.5]: Max value at x=1, $f(1) = 10 - 1^2 = 9$. Area = $9 \times 0.5 = 4.5$4. [1.5, 2]: Max value at x=1.5, $f(1.5) = 10 - (1.5)^2 = 7.75$. Area = $7.75 \times 0.5 = 3.875$5. [2, 2.5]: Max value at x=2, $f(2) = 10 - 2^2 = 6$. Area = $6 \times 0.5 = 3$6. [2.5, 3]: Max value at x=2.5, $f(2.5) = 10 - (2.5)^2 = 3.75$. Area = $3.75 \times 0.5 = 1.875$Summing these areas: $5 + 4.875 + 4.5 + 3.875 + 3 + 1.875 = 23.125$.Therefore, the correct answer is 23.125.

Suppose S6 is the Upper Sum of the Area Enclosed f(x)=10−x2,f ( x ) = 10 - x ^ { 2 },f(x)=10−x2,

Suppose S₆is the Upper Sum of the Area Enclosed $f ( x ) = 10 - x ^ { 2 },$