37. Choose the correct answer:

The area bounded by the curve y = x|x|, axis and the ordinates x = –1 and x = 1 is given by:

[Hint: y = x² if x > 0 and y = –x² if x < 0]

(A) 0

(B) 1/3

(C) 2/3

(D) 4/3

The given curve is $y=x\left|x\right|$

$=\left\{\left(x.xifx\ge 0\right)\left(x.\left(-x\right)ifx\le 0\right)\right\}=\left\{\left(x^{2}if,x\ge 0\right)\left(-x^{2}if,x\le 0\right)\right\}$

Which is in the form of a parabola nad the lines are $x=-1\&x-axis$

At $x=1>0,y=1^2=1$

At $$$x=-1<0,y=-1^2=-1$

Shaded area of the Ist quadrant

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}ydx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

Shaded area of the IInd quadrant

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{0}{\int }}ydx}={\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx}\\ ={\left[\frac{-x^3}{-3}\right]}_{-1}^0\\ =\frac{1}{3}\end{array}$

$\therefore$ Total area of the enclosed region $=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}uni{t}^2$

$\therefore$ Option (c) is correct.

<p>The given curve is&nbsp;<math><mrow><mi>y</mi><mo>=</mo><mi>x</mi><mrow><mo>|</mo><mrow><mi>x</mi></mrow><mo>|</mo></mrow></mrow></math></p><p><math><mrow><mo>=</mo><mrow><mo>{</mo><mrow><mrow><mo>(</mo><mrow><mi>x</mi><mo>.</mo><mi>x</mi><mi>i</mi><mi>f</mi><mi>x</mi><mo>≥</mo><mn>0</mn></mrow><mo>)</mo></mrow><mrow><mo>(</mo><mrow><mi>x</mi><mo>.</mo><mrow><mo>(</mo><mrow><mo>−</mo><mi>x</mi></mrow><mo>)</mo></mrow><mi>i</mi><mi>f</mi><mi>x</mi><mo>≤</mo><mn>0</mn></mrow><mo>)</mo></mrow></mrow><mo>}</mo></mrow><mo>=</mo><mrow><mo>{</mo><mrow><mrow><mo>(</mo><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>i</mi><mi>f</mi><mo>,</mo><mi>x</mi><mo>≥</mo><mn>0</mn></mrow><mo>)</mo></mrow><mrow><mo>(</mo><mrow><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>i</mi><mi>f</mi><mo>,</mo><mi>x</mi><mo>≤</mo><mn>0</mn></mrow><mo>)</mo></mrow></mrow><mo>}</mo></mrow></mrow></math></p><p>Which is in the form of a parabola nad the lines are&nbsp;<math><mrow><mi>x</mi><mo>=</mo><mo>−</mo><mn>1</mn><mo>&amp;</mo><mi>x</mi><mo>−</mo><mi>a</mi><mi>x</mi><mi>i</mi><mi>s</mi></mrow></math></p><p>At&nbsp;<math><mrow><mi>x</mi><mo>=</mo><mn>1</mn><mo>&gt;</mo><mn>0</mn><mo>,</mo><mi>y</mi><mo>=</mo><msup><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>1</mn></mrow></math></p><p>At&nbsp;<math><mrow><mi></mi></mrow></math><strong><math><mrow><mi>x</mi><mo>=</mo><mo>−</mo><mn>1</mn><mo>&lt;</mo><mn>0</mn><mo>,</mo><mi>y</mi><mo>=</mo><mo>−</mo><msup><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mo>−</mo><mn>1</mn></mrow></math></strong></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1742810024phprRvLQS_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1742810024phprRvLQS.jpeg" alt="" width="224" height="198"></picture></div></div><p>Shaded area of the Ist quadrant</p><p><math><mtable columnalign="left"><mtr><mtd><mrow><mo>=</mo><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></munderover><mrow><mi>y</mi><mi>d</mi><mi>x</mi></mrow></mrow></mstyle><mo>=</mo><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></munderover><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>d</mi><mi>x</mi></mrow></mrow></mstyle></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><msubsup><mrow><mrow><mo>[</mo><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mo>]</mo></mrow></mrow><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></msubsup></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mtd></mtr></mtable></math></p><p>Shaded area of the IInd quadrant</p><p><math><mtable columnalign="left"><mtr><mtd><mrow><mo>=</mo><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mo>−</mo><mn>1</mn></mrow><mrow><mn>0</mn></mrow></munderover><mrow><mi>y</mi><mi>d</mi><mi>x</mi></mrow></mrow></mstyle><mo>=</mo><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mo>−</mo><mn>1</mn></mrow><mrow><mn>0</mn></mrow></munderover><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>d</mi><mi>x</mi></mrow></mrow></mstyle></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><msubsup><mrow><mrow><mo>[</mo><mrow><mfrac><mrow><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mo>−</mo><mn>3</mn></mrow></mfrac></mrow><mo>]</mo></mrow></mrow><mrow><mo>−</mo><mn>1</mn></mrow><mrow><mn>0</mn></mrow></msubsup></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mtd></mtr></mtable></math></p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;Total area of the enclosed region&nbsp;<math><mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></math></p><p><math><mrow><mo>=</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>u</mi><mi>n</mi><mi>i</mi><msup><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></math></p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;Option (c) is correct.</p>

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