Question

#JEE Question


@Areas From Calculus




Q):-)The area bounded by
$\red {\tt{y ^ 2 = 4x}}$
with the lines
$\huge{ \pink{x = 2 \: \: \: \: and \: \: \: \: \: x = 5 }}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve are

$\rm :\longmapsto\: {y}^{2} = 4x - - - (1)$

$\rm :\longmapsto\:x = 2 - - - (2)$

and

$\rm :\longmapsto\:x = 5 - - - (3)$

Now,

y² = 4x represents a right handed parabola having vertex at ( 0, 0 ).

Also,

x = 5 is a line parallel to y - axis passes through (5, 0)

Also,

x = 2 is a line parallel to y - axis passes through (2, 0)

So, required area bounded by the curve y² = 4ax between the lines x = 2 and x = 5 is

$\rm \: = \: \: 2\displaystyle \rm \int _2 ^{5} y \: dx$

$\rm \: = \: \: 2\displaystyle \rm \int _2 ^{5} \sqrt{4x} \: dx$

$\rm \: = \: \: 2\displaystyle \rm \int _2 ^{5} 2\sqrt{x} \: dx$

$\rm \: = \: \: 4\displaystyle \rm \int _2 ^{5} \sqrt{x} \: dx$

$\rm \: = \: \: 4\displaystyle \rm \int _2 ^{5} {\bigg(x\bigg) }^{\dfrac{1}{2} } \: dx$

We know that,

$\purple{\boxed{ \bf \: \displaystyle \rm \int {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

$\rm \: = \: \: 4 \: \bigg(\dfrac{{\bigg(x\bigg) }^{\dfrac{1}{2} + 1}}{\dfrac{1}{2} + 1} \bigg)_2 ^{5}$

$\rm \: = \: \: 4 \: \bigg(\dfrac{{\bigg(x\bigg) }^{\dfrac{3}{2}}}{\dfrac{3}{2} } \bigg)_2 ^{5}$

$\rm \: = \: \: \dfrac{8}{3}{\bigg(x\bigg) }^{\dfrac{3}{2} } \bigg |_2 ^{5}$

$\rm \: = \: \: \dfrac{8}{3} \bigg \{{\bigg(5\bigg) }^{\dfrac{3}{2} } - {\bigg(2\bigg) }^{\dfrac{3}{2} } \bigg \}$

$\rm \: = \: \: \dfrac{8}{3} \bigg \{{\bigg( {( \sqrt{5)} }^{2} \bigg) }^{\dfrac{3}{2} } - {\bigg( {( \sqrt{2}) }^{2} \bigg) }^{\dfrac{3}{2} } \bigg \}$

$\rm \: = \: \: \dfrac{8}{3} \bigg \{5 \sqrt{5} - 2 \sqrt{2} \bigg \} \: sq. \: units$

Answer 2

Answer:

⟼y

2

=4x−−−(1)

\rm :\longmapsto\:x = 2 - - - (2):⟼x=2−−−(2)

and

\rm :\longmapsto\:x = 5 - - - (3):⟼x=5−−−(3)

Now,

y² = 4x represents a right handed parabola having vertex at ( 0, 0 ).

Also,

x = 5 is a line parallel to y - axis passes through (5, 0)

Also,

x = 2 is a line parallel to y - axis passes through (2, 0)

So, required area bounded by the curve y² = 4ax between the lines x = 2 and x = 5 is

\rm \: = \: \: 2\displaystyle \rm \int _2 ^{5} y \: dx=2∫

2

5

ydx

\rm \: = \: \: 2\displaystyle \rm \int _2 ^{5} \sqrt{4x} \: dx=2∫

2

5

4x

dx

\rm \: = \: \: 2\displaystyle \rm \int _2 ^{5} 2\sqrt{x} \: dx=2∫

2

5

2

x

dx

\rm \: = \: \: 4\displaystyle \rm \int _2 ^{5} \sqrt{x} \: dx=4∫

2

5

x

dx

\rm \: = \: \: 4\displaystyle \rm \int _2 ^{5} {\bigg(x\bigg) }^{\dfrac{1}{2} } \: dx=4∫

2

5

(x)

2

1

dx

We know that,

\purple{\boxed{ \bf \: \displaystyle \rm \int {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c}}

∫x

n

dx=

n+1

x

n+1

+c

\rm \: = \: \: 4 \: \bigg(\dfrac{{\bigg(x\bigg) }^{\dfrac{1}{2} + 1}}{\dfrac{1}{2} + 1} \bigg)_2 ^{5}=4(

2

1

+1

(x)

2

1

+1

)

2

5

\rm \: = \: \: 4 \: \bigg(\dfrac{{\bigg(x\bigg) }^{\dfrac{3}{2}}}{\dfrac{3}{2} } \bigg)_2 ^{5}=4(

2

3

(x)

2

3

)

2

5

\rm \: = \: \: \dfrac{8}{3}{\bigg(x\bigg) }^{\dfrac{3}{2} } \bigg |_2 ^{5}=

3

8

(x)

2

3

∣

∣

∣

∣

∣

2

5

\rm \: = \: \: \dfrac{8}{3} \bigg \{{\bigg(5\bigg) }^{\dfrac{3}{2} } - {\bigg(2\bigg) }^{\dfrac{3}{2} } \bigg \}=

3

8

{(5)

2

3

−(2)

2

3

}

\rm \: = \: \: \dfrac{8}{3} \bigg \{{\bigg( {( \sqrt{5)} }^{2} \bigg) }^{\dfrac{3}{2} } - {\bigg( {( \sqrt{2}) }^{2} \bigg) }^{\dfrac{3}{2} } \bigg \}=

3

8

{((

5)

2

)

2

3

−((

2

)

2

)

2

3

}

\rm \: = \: \: \dfrac{8}{3} \bigg \{5 \sqrt{5} - 2 \sqrt{2} \bigg \} \: sq. \: units=

3

8

{5

5

−2

2

}sq.units

Step-by-step explanation:

$\fbox\red{Thank you}$