Suppose represents the velocity of an object on an interval .

Using the Riemann sum approximation, , to approximate the area under the curve estimates which of the following?

Question

Suppose   represents the velocity of an object on an interval .

Using the Riemann sum approximation,    , to approximate the area under the curve  estimates which of the following?

BlackTom AI · Accepted Answer

Reframing the question in my own words: we are looking at the area under the velocity curve v(t) = f(x) over the interval [a, b] and asking what the Riemann-sum-based approximation of that area estimates.
Option presented: 'the total distance traveled by the object over the interval [a,b].' 
First, consider the fundamental relationship: the integral of velocity v(t) with respect to time gives the net displacement, which is the change in position, not necessarily the total distance traveled.
If the velocity v(t) is nonnegative for all t in [a, b], then the area under the velocity curve equals the distance traveled, because all motion contributes positively to distance. In that special case, the area and the distance coincide.
However, if v(t) takes negative values on any subinterval of [a, b], then the area under the curve subtracts the negative portions, yielding net displacement rather than total distance. The total distance traveled would instead be the integral of the absolute value |v(t)|, which accounts for all movement without cancellation.
Given the generic statement about using the area under the velocity curve to estimate something, the standard interpretation is that the area estimates displacement, not necessarily the total distance unless you know v(t) ≥ 0 on the entire interval. This means the option claiming it estimates total distance traveled is not universally correct; it would only be correct if velocity never goes negative on [a, b].
To summarize the key idea in a different way: the Riemann sum for ∫_a^b v(t) dt approximates the net change in position, i.e., displacement. Distance traveled, in general, requires ∫_a^b |v(t)| dt, which is a different integral. The graph’s behavior (whether v(t) stays above the axis or not) determines which interpretation applies in a given situation.

MATH_1225_64262_202506 5.1 Areas and Distances

Suppose represents the velocity of an object on an interval . Using the Riemann sum approximation, , to approximate the area under the curve estimates which of the following?

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MATH_1225_64262_202506 5.1 Areas and Distances

Suppose represents the velocity of an object on an interval . Using the Riemann sum approximation, , to approximate the area under the curve estimates which of the following?

View Explanation

Log in for full answers

Similar Questions

Match the type of estimation technique shown in the pictures with the correct label. 1: ____ 2: ____ 3: ____ 4: ____

Find the difference between the upper and lower estimates of the distance traveled at velocity 25−t2 on the interval 1≤t≤4 for 1000 subdivisions.

Express the following integral as a limit of Riemann sums:

Suppose we want to find the area under the curve (pictured below) on the interval using a Riemann sum with rectangles. Which of the following provides the most accurate approximation of the area under the curve on the given interval?

Suppose we want to find the area under the curve (pictured below) on the interval using a Riemann sum with rectangles. Which of the following provides the most accurate approximation of the area under the curve on the given interval?

You should see 2 equations in this question. Does the 2nd equation render correctly in your browser? It should look the same as the image in this box: ∫ 0 1 𝑥 2 𝑑 𝑥 = lim 1 𝑛 ( 1 2 𝑛 2 + 2 2 𝑛 2 + . . . + ( 𝑛 2 𝑛 2 ) ) = lim ∑ 𝑛 𝑚 𝑥 × 𝑛 2 2 3

You should see 2 equations in this question. Does the 2nd equation render correctly in your browser? It should look the same as the image in this box: ∫ 1 0 x2dx=lim 1 n ( 12 n2 + 22 n2 +...+( n2 n2 ))=lim∑ n m x × n2 23

More Practical Tools for Students Powered by AI Study Helper

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Match the type of estimation technique shown in the pictures with the correct label. 1: 2: 3: 4: