Question at position 8 The exact area of the region bounded by the graphs of y = x and y = x2 is1/3 sq unit.1/6 sq unit.1/2 sq unit.2/3 sq unit.5/6 sq unit.单项选择题
A
1/3 sq unit.
B
1/6 sq unit.
C
1/2 sq unit.
D
2/3 sq unit.
E
5/6 sq unit.

Question

Question at position 8 The exact area of the region bounded by the graphs of y = x and y = x2 is1/3 sq unit.1/6 sq unit.1/2 sq unit.2/3 sq unit.5/6 sq unit.

BlackTom AI · Accepted Answer

Question restatement: We are asked to find the area of the region bounded by the graphs y = x and y = x^2, between their intersection points.
Option 1: 1/3 sq unit. If one incorrectly takes the area as ∫0^1 (x) dx or uses an improper top function, you might get 1/3, but the correct setup is to integrate the difference between the two curves over the interval where they enclose a region. Here, x and x^2 intersect at x = 0 and x = 1, so the area would be ∫ from 0 to 1 of (top − bottom) dx. 1/3 would arise from a mistaken integrand or limits.
Option 2: 1/6 sq unit. This matches the standard calculation. Over 0 ≤ x ≤ 1, x ≥ x^2, so the vertical distance is x − x^2. Integrating: ∫0^1 (x − x^2) dx = [x^2/2 − x^3/3]0^1 = (1/2 − 1/3) = 1/6. This is the correct area for the bounded region.
Option 3: 1/2 sq unit. This would correspond to misevaluating the integral as ∫0^1 x dx, which ignores the x^2 term, or misidentifying which function is on top. Since x^2 is not the top function over the entire interval [0,1], this is not the area of the region between the curves.
Option 4: 2/3 sq unit. This result would come from neither the proper limits nor the proper integrand; it’s larger than the total area under y = x over [0,1] and indicates a conceptual error in subtracting the lower curve from the upper curve. The correct difference integrates to less than 1/2, so 2/3 is too big.
Option 5: 5/6 sq unit. This is even more inconsistent with the actual geometry. The bounded region lies between x = 0 and x = 1 and has area far smaller than 1, since the integrand x − x^2 is at most 1/4 at x = 1/2 and the interval length is 1. A result of 5/6 would require a gross overcounting or incorrect sign or limits, which does not reflect the true area.

Question at position 8 The exact area of the region bounded by the graphs of y = x and y = x2 is1/3 sq unit.1/6 sq unit.1/2 sq unit.2/3 sq unit.5/6 sq unit.单项选择题

类似问题

Which one of the following is equal to the shaded area above?

The area bounded by the two graphs above is equal to:

Find the shaded area above.

Which one of the following is equal to the shaded area above?

Find the exact shaded area above.

Find the area of the region bounded by the graphs of the function , , . Round your answer to three decimal places.

Find the area of the region bounded by the graphs of the algebraic functions.

Find the area of the region bounded by the equations by integrating (i) with respect to x and (ii) with respect to y.

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单

Question at position 8 The exact area of the region bounded by the graphs of y = x and y = x2 is1/3 sq unit.1/6 sq unit.1/2 sq unit.2/3 sq unit.5/6 sq unit.单项选择题

类似问题

Which one of the following is equal to the shaded area above?

The area bounded by the two graphs above is equal to:

Find the shaded area above.

Which one of the following is equal to the shaded area above?

Find the exact shaded area above.

Find the area of the region bounded by the graphs of the function , , . Round your answer to three decimal places. ​

Find the area of the region bounded by the graphs of the algebraic functions. ​ ​

Find the area of the region bounded by the equations by integrating (i) with respect to x and (ii) with respect to y. ​ ​

更多留学生实用工具

希望你的学习变得更简单

Find the area of the region bounded by the graphs of the function , , . Round your answer to three decimal places.

Find the area of the region bounded by the graphs of the algebraic functions.

Find the area of the region bounded by the equations by integrating (i) with respect to x and (ii) with respect to y.