Q.10 Evaluate: ∫₋₁¹ |x| dx

Answer: 1 |x| = -x for x lt 0 and |x| = x for x ≥ 0 . ∫₋₁¹|x| dx = ∫₋₁⁰(-x) dx + ∫₀¹x dx = [-x²/2]₋₁⁰ + [x²/2]₀¹ = 1/2 + 1/2 = 1 . (Alternatively, |x| is even, so the integral is 2∫₀¹ x dx = 1 .)

Evaluate: \(\int_{-1}^{1} |x|\,dx\)

Q.10 Evaluate: \(\int_{-1}^{1} |x|\,dx\)

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: 1

\(|x| = -x\) for \(x \lt 0\) and \(|x| = x\) for \(x \ge 0\).

\(\int_{-1}^{1}|x|\,dx = \int_{-1}^{0}(-x)\,dx + \int_{0}^{1}x\,dx = \left[-\frac{x^2}{2}\right]_{-1}^{0} + \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2} + \frac{1}{2} = 1\).

(Alternatively, \(|x|\) is even, so the integral is \(2\int_0^1 x\,dx = 1\).)

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Evaluate: \(\int_{-1}^{1} |x|\,dx\)

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