Antiderivative Of Sqrt 9 X 2: The Setup Makes All The Difference

Last Updated: Written by Prof. Daniel Marques de Lima
antiderivative of sqrt 9 x 2 the setup makes all the difference
antiderivative of sqrt 9 x 2 the setup makes all the difference
Table of Contents

Antiderivative of sqrt 9 x 2 Feels Harder Than It Is

The antiderivative of sqrt(9x^2) is not as complicated as it might appear. Since sqrt(9x^2) = 3|x|, the indefinite integral splits into two simple cases depending on the sign of x. For x ≥ 0, the integrand is 3x, and for x < 0, it is -3x. Recognizing this piecewise structure lets us write a concise antiderivative that captures all x.

Concretely, the integral ∫ sqrt(9x^2) dx simplifies to ∫ 3|x| dx. Using the standard antiderivative rules, we obtain different expressions on each interval, then combine them with an explicit piecewise definition. The final, general antiderivative can be written as F(x) = (3/2)x|x| + C, which automatically handles both signs of x. This form reflects the geometry of the area under the absolute value function and aligns with the Marist educational emphasis on clarity and rigor.

Why the Piecewise View Helps

Interpreting sqrt(9x^2) as 3|x| clarifies the underlying symmetry around x = 0. The derivative of (1.5)x|x| is 3|x| for all x ≠ 0, matching the integrand. At x = 0, the derivative is not defined in the classical sense, but the antiderivative remains valid in the sense of an indefinite integral with a constant of integration. This approach avoids misapplying the rule sqrt(a^2) = a when a may be negative.

Step-by-Step Derivation

  1. Rewrite the integrand: sqrt(9x^2) = 3|x|.
  2. Split the domain at x = 0 to handle the absolute value: - For x ≥ 0, |x| = x, so ∫3x dx = (3/2)x^2 + C. - For x < 0, |x| = -x, so ∫(-3x) dx = -(3/2)x^2 + C.
  3. Combine into a single expression: F(x) = (3/2)x|x| + C.

Common Variations

Depending on context, you might see the antiderivative presented with a domain-specific constant or piecewise form. Here are two equivalent representations:

  • F(x) = (3/2)x^2 for x ≥ 0; F(x) = -(3/2)x^2 for x < 0; plus a single constant C across the domain.
  • F(x) = (3/2)x|x| + C (compact form), valid for all real x as an antiderivative in the Lebesgue sense.
antiderivative of sqrt 9 x 2 the setup makes all the difference
antiderivative of sqrt 9 x 2 the setup makes all the difference

Practical Check

Differentiate F(x) = (3/2)x|x| to verify: for x > 0, F'(x) = 3x; for x < 0, F'(x) = -3x; at x = 0 the derivative from the left and right are 0, aligning with the integrand's behavior almost everywhere. This supports F as a correct antiderivative in the standard calculus framework used in advanced Marist pedagogy.

Concrete Example

Compute the antiderivative of sqrt(9x^2) over an interval, say from x = -2 to x = 3. Using the compact form, the definite integral would be F - F(-2) with F(x) = (3/2)x|x|. Evaluate: F = (3/2)(3) = 13.5; F(-2) = (3/2)(-2) = -6. So the integral equals 13.5 - (-6) = 19.5. This concrete result illustrates how the absolute value structure contributes asymmetrically to areas on opposite sides of the origin.

Key Takeaways

- sqrt(9x^2) simplifies to 3|x|, not 3x always. Structural clarity comes from acknowledging absolute value.

- The antiderivative is neatly captured by F(x) = (3/2)x|x| + C, which encodes both sign regimes in one formula.

FAQ

CaseIntegrandAntiderivative
x ≥ 03x(3/2)x^2 + C
x < 0-3x-(3/2)x^2 + C
Combined3|x|(3/2)x|x| + C

Note: In educational contexts aligned with Marist pedagogy, this derivation reinforces foundational calculus concepts while respecting the value-based emphasis on clarity, rigor, and accessibility for diverse learners across Brazil and Latin America.

Helpful tips and tricks for Antiderivative Of Sqrt 9 X 2 The Setup Makes All The Difference

[Question]?

[Answer] The question is addressed in the main text. The antiderivative is F(x) = (3/2)x|x| + C, derived from recognizing sqrt(9x^2) = 3|x| and integrating piecewise over x ≥ 0 and x < 0.

[Question]?

[Answer] For x ≥ 0, the integral of sqrt(9x^2) dx is (3/2)x^2 + C. For x < 0, it is -(3/2)x^2 + C. A single compact form that covers both is F(x) = (3/2)x|x| + C.

[Question]?

[Answer] Why does the absolute value matter here? Because squaring x to obtain sqrt(x^2) hides the sign of x; using 3|x| preserves the correct algebraic behavior and leads to a correct, globally valid antiderivative when combined with the constant of integration.

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Prof. Daniel Marques de Lima

Prof. Daniel Marques de Lima is a veteran educator-researcher with 25 years in university-affiliated teacher preparation programs and Marist school networks across Brazil.

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