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The definite integral of √(x³) over the limits 0 to 2 is

A) ∫(0 to 2) √(x³) dx = 3
B) ∫(0 to 2) √(x³) dx = 4
C) ∫(0 to 2) √(x³) dx = 2
D) ∫(0 to 2) √(x³) dx = 1

User Ngg
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1 Answer

4 votes

Final answer:

The definite integral of √(x³) over the limits 0 to 2 is 3.2, which is not listed among the provided options. This is found by integrating x^(3/2) and evaluating the antiderivative at the upper and lower limits.

Step-by-step explanation:

The question concerns the computation of a definite integral of the square root of x cubed (√(x³)) over the interval from 0 to 2. We can express the integral as ∫(√(x³) dx) from 0 to 2. First, we rewrite the integrand √(x³) as x^(3/2) to simplify the integral. Now, we'll integrate x^(3/2) with respect to x over the given limits:

  1. Express the integral in terms of x: ∫(0 to 2) x^(3/2) dx.
  2. Find the antiderivative of x^(3/2), which is (2/5)x^(5/2).
  3. Evaluate the antiderivative at the upper and lower limits and subtract: [(2/5)*(2)^(5/2)] - [(2/5)*(0)^(5/2)].
  4. Simplify the result to obtain the value of the integral.

The calculation yields the value:

((2/5)*(2)^(5/2)) − ((2/5)*(0)^(5/2)) = 3.2.

Thus, none of the provided options A) 3, B) 4, C) 2, or D) 1 are correct. The exact value is 3.2, which is not listed among the answer choices.