The integral of (\sqrt{4x^{2}}) is a classic example that combines algebraic simplification with careful attention to the absolute value function, making it an excellent teaching tool for students learning how to handle radicals and piecewise expressions in calculus. By working through this problem, learners reinforce their understanding of square roots, the properties of absolute value, and the techniques for finding both indefinite and definite integrals. The process also highlights why Consider the domain of the integrand before applying standard integration rules — this one isn't optional.
Understanding the Integrand
The expression under the integral sign is (\sqrt{4x^{2}}). That said, this shortcut overlooks a crucial detail: the square root function always returns a non‑negative result. So, (\sqrt{4x^{2}}) is actually equal to (2|x|), where (|x|) denotes the absolute value of (x). In practice, at first glance, one might be tempted to cancel the square root and the square, writing (\sqrt{4x^{2}} = 2x). Recognizing this distinction is the first step toward a correct evaluation.
Why the Absolute Value Appears
Recall the definition of the square root for real numbers: (\sqrt{a^{2}} = |a|) for any real (a). Applying this rule with (a = 2x) gives
[ \sqrt{(2x)^{2}} = |2x| = 2|x|. ]
If we ignored the absolute value and wrote (2x) outright, the resulting antiderivative would be correct only on intervals where (x \ge 0). On intervals where (x < 0), the sign would be wrong, leading to errors in definite integrals that cross zero The details matter here. Still holds up..
Simplifying the Integral
With the integrand rewritten as (2|x|), the indefinite integral becomes
[ \int \sqrt{4x^{2}} , dx = \int 2|x| , dx. ]
Because the absolute value creates a piecewise definition, we treat the integral separately for (x \ge 0) and (x < 0).
Piecewise Integration
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For (x \ge 0): (|x| = x), so
[ \int 2|x| , dx = \int 2x , dx = x^{2} + C_{1}. ]
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For (x < 0): (|x| = -x), so
[ \int 2|x| , dx = \int -2x , dx = -x^{2} + C_{2}. ]
These two expressions can be combined into a single formula using the absolute value itself:
[ \int 2|x| , dx = x|x| + C, ]
where (C) is an arbitrary constant. To verify, differentiate (x|x|):
[ \frac{d}{dx}\bigl(x|x|\bigr) = |x| + x \cdot \frac{d}{dx}|x| = |x| + x \cdot \operatorname{sgn}(x) = |x| + |x| = 2|x|, ]
which matches the integrand. Hence, the indefinite integral of (\sqrt{4x^{2}}) is
[ \boxed{\int \sqrt{4x^{2}} , dx = x|x| + C}. ]
Working with Definite Integrals
When evaluating a definite integral (\int_{a}^{b} \sqrt{4x^{2}} , dx), the piecewise nature of the integrand often requires splitting the interval at any point where the expression inside the absolute value changes sign—namely, at (x = 0) Simple, but easy to overlook. Still holds up..
Example 1: Both Limits Positive
Compute (\int_{1}^{3} \sqrt{4x^{2}} , dx). Since the interval ([1,3]) lies entirely in the region (x \ge 0), we can replace (\sqrt{4x^{2}}) with (2x):
[ \int_{1}^{3} 2x , dx = \bigl[x^{2}\bigr]_{1}^{3} = 3^{2} - 1^{2} = 9 - 1 = 8. ]
Example 2: Limits Spanning Zero
Evaluate (\int_{-2}^{2} \sqrt{4x^{2}} , dx). Here the interval crosses zero, so we split it:
[ \int_{-2}^{2} 2|x| , dx = \int_{-2}^{0} 2(-x) , dx + \int_{0}^{2} 2x , dx. ]
Compute each part:
[ \int_{-2}^{0} -2x , dx = \bigl[-x^{2}\bigr]{-2}^{0} = -(0^{2}) + (-(-2)^{2}) = 0 - 4 = -4, ] [ \int{0}^{2} 2x , dx = \bigl[x^{2}\bigr]_{0}^{2} = 4 - 0 = 4. ]
Adding them gives (-4 + 4 = 0). The result makes sense geometrically: the function (2|x|) is symmetric about the y‑axis, and the signed area over ([-2,0]) cancels the area over ([0,2]).
Example 3: Both Limits Negative
For (\int_{-5}^{-1} \sqrt{4x^{2}} , dx), the entire interval lies where (x < 0
