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How do you evaluate #int (x^2 - 2x)dx# for [0, 1]?

Answer 1

Emily Ammerman

#-2/3#

#int_0^1(x^2-2x)dx=[x^3/3-(2x^2)/2]_0^1#

#=1/3-1=-2/3#

(Note that this will represent the area under the graph of #y=x^2-2x# between the points #x=0 and x=1#)

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Answer 2

Evelyn Agard

To evaluate ∫(x^2 - 2x)dx for the interval [0, 1], follow these steps:

Integrate the expression (x^2 - 2x) with respect to x.
Evaluate the result at the upper limit (1) and subtract from it the result evaluated at the lower limit (0).

Step 1: ∫(x^2 - 2x)dx = (1/3)x^3 - x^2

Step 2: Evaluate the expression at x = 1: (1/3)(1)^3 - (1)^2 = 1/3 - 1 = -2/3

Evaluate the expression at x = 0: (1/3)(0)^3 - (0)^2 = 0

Step 3: Subtract the lower limit result from the upper limit result: (-2/3) - 0 = -2/3

So, the value of ∫(x^2 - 2x)dx for the interval [0, 1] is -2/3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.