Interpret the definite integral A = integral [a,b] dA = integral L(x)dx as computing the area of a region in the xy plane. Then one can think of the definite integral as:

A. "accumulating" all of the small segments of area "dA" from a to b
B. the antiderivative of L(x)
C. "accumulating" all of the small segments of area "dA" from a to b AND "accumulating" all of the small segments of area "L(x) dx" from a to b, where L(x) represents the length of a rectangle at a particular x value, and dx the width
D. the antiderivative of dA
E. "accumulating" all of the small segments of area "L(x) dx" from a to b, where L(x)r epresents the length of a rectangle at a particular x value, and dx the width

I like C the most

The correct interpretation of the definite integral as computing the area of a region in the xy plane is option C.

Option A, "accumulating" all of the small segments of area "dA" from a to b, is a part of the process in finding the total accumulated area using the definite integral. The integral sums up all the infinitesimally small segments of area "dA" over the interval from a to b.

Option B, the antiderivative of L(x), is not the correct interpretation. While the indefinite integral (without limits) does calculate the antiderivative of a function, the definite integral computes the area under the curve between the specified limits a and b.

Option D, the antiderivative of dA, is similar to option B and is not the correct interpretation. Again, the definite integral is used for calculating the area, not the antiderivative.

Option E, "accumulating" all of the small segments of area "L(x) dx" from a to b, is partly correct. However, it only represents the width (dx) multiplied by the length (L(x)) of each infinitesimally small rectangle. Whereas, option C includes both the accumulation of small segments of area "dA" (width times length) and the accumulation of the small segments of area "L(x) dx" (length times width).

Therefore, the correct interpretation is option C: "accumulating" all of the small segments of area "dA" from a to b AND "accumulating" all of the small segments of area "L(x) dx" from a to b, where L(x) represents the length of a rectangle at a particular x value, and dx the width.

Answer is B (Apex)

E. adding height L times width dx at x = a, at x = a + dx, at x = a+2dx ....to x = b (with tiny, like vanishing, dx)