20. Choose the correct answer:

Area lying between the curves y² = 4x and y = 2x is

(A) 2/3

(B) 1/3

(C) 1/4

(D) 3/4

The given equation of the curve is $y^2=4x$ - (1) and

the line is $y=2x$ - (2)

Solving (1) and (2) for x and y

$\begin{array}{l}{\left(2x\right)}^2=4x\\ =4x^2=4x\\ =x^2-x=0\\ =x\left(x-1\right)=0\end{array}$

So,  $x=0\&x=1$

for $x=0$ we get $y=2\times 0=0$

for $x=1$ , we get $y=2\times 1=2$

so, the point of intersection are (0,0)and (1,2)

area (DCAO)=area (DCABO)-area ( $?OAB$ )

<p>The given equation of the curve is&nbsp;<math><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>4</mn><mi>x</mi></mrow></math>&nbsp;- (1) and</p><p>the line is&nbsp;<math><mrow><mi>y</mi><mo>=</mo><mn>2</mn><mi>x</mi></mrow></math>&nbsp;- (2)</p><p>Solving (1) and (2) for x and y</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1742801480phpi3FBIe_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1742801480phpi3FBIe.jpeg" alt="" width="306" height="250"></picture></div><div><span contenteditable="false"> <math> <mtable columnalign="left"> <mtr> <mtd> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>4</mn> <mi>x</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mn>4</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>4</mn> <mi>x</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mi>x</mi> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mi>x</mi> <mrow> <mo> (</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </math> </span></div></div><p>So, &nbsp;<math><mrow><mi>x</mi><mo>=</mo><mn>0</mn><mo>&amp;</mo><mi>x</mi><mo>=</mo><mn>1</mn></mrow></math></p><p>for&nbsp;<math><mrow><mi>x</mi><mo>=</mo><mn>0</mn></mrow></math>&nbsp;we get&nbsp;<math><mrow><mi>y</mi><mo>=</mo><mn>2</mn><mo>×</mo><mn>0</mn><mo>=</mo><mn>0</mn></mrow></math></p><p>for&nbsp;<math><mrow><mi>x</mi><mo>=</mo><mn>1</mn></mrow></math>&nbsp;, we get&nbsp;<math><mrow><mi>y</mi><mo>=</mo><mn>2</mn><mo>×</mo><mn>1</mn><mo>=</mo><mn>2</mn></mrow></math></p><p>so, the point of intersection are (0,0)and (1,2)</p><p>area (DCAO)=area (DCABO)-area (&nbsp;<math><mrow><mo>? </mo><mi>O</mi><mi>A</mi><mi>B</mi></mrow></math>&nbsp;)</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1742801555phpbQVEE4_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1742801555phpbQVEE4.jpeg" alt="" width="241" height="390"></picture></div></div>

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