31. Using the method of integration, find the area enclosed by the curve |x| + |y| = 1

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]

Given equation of the curve is $\left|x\right|+\left|y\right|=1$ , which can be break down into each quadrant .

For Ist quadrant,  $\left|x\right|=x,\left|y\right|=y$

i.e.,  $x+y=1$ - (1)

Similarly for IInd, IIIRd nad IVth quadrant

$-x+y=1$ - (2)

$-x-y=1$ - (3)

$x-y=1$ - (4)

We draw the above focus lines on a graph and find the area enclosed which is a square.

$\therefore$ Required area $\left(?ABCD\right)=4\times area\left(?AOB\right)$ .

$\begin{array}{l}=4\times {\displaystyle \underset{0}{\overset{1}{\int }}y}dx\\ =4{\displaystyle \underset{0}{\overset{1}{\int }}\left(-x\right)}dx\\ =4{\left[x-\frac{x^2}{2}\right]}_0^1\\ =4\left[1-\frac{1}{2}\right]\\ =4\times \frac{1}{2}\\ =2uni{t}^2\end{array}$

<p>Given equation of the curve is&nbsp;<math><mrow><mo>|</mo><mi>x</mi><mo>|</mo><mo>+</mo><mo>|</mo><mi>y</mi><mo>|</mo><mo>=</mo><mn>1</mn></mrow></math>&nbsp;, which can be break down into each quadrant .</p><p>For Ist quadrant, &nbsp;<math><mrow><mo>|</mo><mi>x</mi><mo>|</mo><mo>=</mo><mi>x</mi><mo>, </mo><mo>|</mo><mi>y</mi><mo>|</mo><mo>=</mo><mi>y</mi></mrow></math></p><p>i.e., &nbsp;<math><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>=</mo><mn>1</mn></mrow></math>&nbsp;- (1)</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1742808962php4CTsJg_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1742808962php4CTsJg.jpeg" alt="" width="221" height="183"></picture></div></div><p>Similarly for IInd, IIIRd nad IVth quadrant</p><p><math><mrow><mo>−</mo><mi>x</mi><mo>+</mo><mi>y</mi><mo>=</mo><mn>1</mn></mrow></math>&nbsp;- (2)</p><p><math><mrow><mo>−</mo><mi>x</mi><mo>−</mo><mi>y</mi><mo>=</mo><mn>1</mn></mrow></math>&nbsp;- (3)</p><p><math><mrow><mi>x</mi><mo>−</mo><mi>y</mi><mo>=</mo><mn>1</mn></mrow></math>&nbsp;- (4)</p><p>We draw the above focus lines on a graph and find the area enclosed which is a square.</p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;Required area&nbsp;<math><mrow><mrow><mo> (</mo><mrow><mo>? </mo><mi>A</mi><mi>B</mi><mi>C</mi><mi>D</mi></mrow><mo>)</mo></mrow><mo>=</mo><mn>4</mn><mo>×</mo><mi>a</mi><mi>r</mi><mi>e</mi><mi>a</mi><mrow><mo> (</mo><mrow><mo>? </mo><mi>A</mi><mi>O</mi><mi>B</mi></mrow><mo>)</mo></mrow></mrow></math>&nbsp;.</p><p><math><mtable columnalign="left"><mtr><mtd><mrow><mo>=</mo><mn>4</mn><mo>×</mo><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></munderover><mrow><mi>y</mi></mrow></mrow></mstyle><mi>d</mi><mi>x</mi></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><mn>4</mn><mstyle displaystyle="true"><mrow><munderover><mo>∫</mo><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></munderover><mrow><mrow><mo> (</mo><mrow><mo>−</mo><mi>x</mi></mrow><mo>)</mo></mrow></mrow></mrow></mstyle><mi>d</mi><mi>x</mi></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><mn>4</mn><msubsup><mrow><mrow><mo> [</mo><mrow><mi>x</mi><mo>−</mo><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>2</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><mn>4</mn><mrow><mo> [</mo><mrow><mn>1</mn><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>]</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><mn>4</mn><mo>×</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></mtd></mtr><mtr><mtd><mrow><mo>=</mo><mn>2</mn><mi>u</mi><mi>n</mi><mi>i</mi><msup><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mtd></mtr></mtable></math></p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment