Q10
1 markMCQSection A

The area bounded by the curve y=x∣x∣y = x|x|, x-axis and the ordinates x=−1 and x=1x = -1 \text{ and } x = 1 is given by

Application of Integrals
Area Bounded by a Modulus Curve

Options

(A)0
(B)13\frac{1}{3}
(C)23\frac{2}{3}
(D)43\frac{4}{3}
Official Answer

Correct option: (C) 23\frac{2}{3}


Since y=x∣x∣y = x|x| is an odd function, the region below the x-axis for x∈[−1,0]x \in [-1,0] mirrors the region above the x-axis for x∈[0,1]x \in [0,1]; the total (unsigned) area is twice the area of one piece: 2×∫01x2 dx=2×(13)2\times \int_0^1 x^2\,dx = 2\times\left(\frac{1}{3}\right) = 23\frac{2}{3}.

area bounded by curvemodulus functionx|x|application of integralssigned vs unsigned areaodd function area

Marking Scheme

  • 11 mark: awarded only for selecting option (C) 2/3; no partial credit.

Hint

Area must be computed as the sum of absolute values of the integral over each region where the curve is on one side of the x-axis — don't just integrate straight through.

Quick Oral Answer

Area equals the sum of the absolute areas on each side of x=0x=0: 13+13=23\frac{1}{3}+\frac{1}{3} = \frac{2}{3}, not the signed integral which wrongly gives 0.

Analysis & Explanation

Concept: Area (not the signed integral) under a piecewise curve.


For x≥0,y=x∣x∣=x2x \ge 0, y = x|x| = x^2. For x<0,y=x∣x∣=−x2x < 0, y = x|x| = -x^2, which lies below the x-axis. Area, being always non-negative, requires taking the absolute value of the curve on each sub-interval.


  • Area from x=−1 to 0=∫∣−x2∣ dx=∫x2 dx=13\text{Area from } x=-1 \text{ to } 0 = \int |-x^2|\,dx = \int x^2\,dx = \frac{1}{3}
  • Area from x=0 to 1=∫x2 dx=13\text{Area from } x=0 \text{ to } 1 = \int x^2\,dx = \frac{1}{3}
  • Total area=13+13=23\text{Total area} = \frac{1}{3}+\frac{1}{3} = \frac{2}{3}

Why the distractors fail:

  • (A) 0 results from wrongly computing the signed definite integral of x|x| directly (which is 0 because x|x| is odd) and mistaking that for the geometric area — a very common conceptual trap in 'area bounded by curve' questions.
  • (B) 1/3 accounts for only one of the two symmetric pieces (e.g. only x in [0,1]), forgetting the mirror region on the negative side.
  • (D) 4/3 typically comes from doubling incorrectly (e.g. using 4×134 \times \frac{1}{3} by miscounting sub-regions).

Exam tip: Whenever a question says 'area bounded by curve and x-axis,' always split at the x-axis crossings and take the absolute value of the integral on each piece — never integrate straight through when the curve changes sign.

Common Mistakes

  1. 1Computing the plain definite integral of x∣x∣ from −1 to 1x|x| \text{ from } -1 \text{ to } 1 (=0) and mistaking it for the area, ignoring that area must always be non-negative.
  2. 2Forgetting to split the interval at x=0x=0 where the sign of y=x∣x∣y = x|x| changes.
  3. 3Only computing area for one half of the interval and doubling incorrectly instead of correctly summing both symmetric halves.

Interesting Facts

This 'area under y=x|x|' style question has appeared repeatedly in CBSE Class 12 board papers and NCERT Exemplar because it tests the frequently-confused distinction between a definite integral's signed value and true geometric area.

The curve y=x∣x∣y = x|x| is actually a smooth (differentiable) curve despite involving |x|, because its derivative 2∣x∣2|x| is continuous everywhere, including at x=0x = 0.

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Frequently Asked Questions

Why isn't the area 0, since the plain integral of x∣x∣ from −1 to 1x|x| \text{ from } -1 \text{ to } 1 is 0?

Because a definite integral gives the signed (algebraic) sum of areas above and below the x-axis, while 'area bounded by the curve' always means the actual non-negative geometric area — so each piece must be taken as a positive quantity before adding.

How do you know where to split the interval?

Split at points where the curve crosses the x-axis (y=0y=0), here at x=0x=0, since the sign of x∣x∣x|x| changes there.