Question

find the area of the region bounded by the curve y²=
x and the lines x=1, x=4
and x axis in 1st quadrant ​

Answer 1

EXPLANATION.

Area of the region bounded by the curves,

y² = x.

the lines x = 1, and x = 4 and the x-axis in 1st quadrant.

As we know that,

The curves y² = x is indicate that the curve is a parabola.

$\sf \implies \int\limits^4_1 {y} \, dx$

From equation,

y² = x.

y = √x.

Put the value of y = √x in equation, we get.

$\sf \implies \int\limits^4_1 {\sqrt{x} } \, dx$

Using the formula,

∫xⁿdx = xⁿ⁺¹/n + 1.

$\sf \implies \bigg[\dfrac{x^{\dfrac{1}{2}+1 } }{\dfrac{1}{2}+1 } \bigg]_1^4$

$\sf \implies \bigg[\dfrac{x^{\dfrac{3}{2} } }{\dfrac{3}{2} } \bigg]_1^4$

$\sf \implies \dfrac{2}{3} \bigg[(x)^{\dfrac{3}{2} } \bigg]_1^4$

$\sf \implies \dfrac{2}{3} \bigg[(4)^{\dfrac{3}{2}} - (1)^{\dfrac{3}{2} } \bigg]$

$\sf \implies\dfrac{2}{3} \bigg[(2)^{3} - (1)^{} \bigg]$

$\sf \implies \dfrac{2}{3} \bigg[8 - 1\bigg]$

$\sf \implies \dfrac{14}{3} units$

                                                                                             

MORE INFORMATION.

(1) = The area bounded by a cartesian curve y = f(x), x-axis and abscissa x = a  and  x = b is given by,

$\sf \implies Area = \int\limits^b_a {y} \, dx =\int\limits^b_a {f(x)} \, dx$

(2) = The area bounded  by a cartesian curve x = f(y), y-axis and ordinates y = c and y = d.

$\sf \implies Area = \int\limits^d_c {x} \, dy = \int\limits^d_c {f(y)} \, dy$

Answer 2

✅ See the attachment diagram.

Lᴇᴛ,

• ABOCD represent the curve y² = x (Parabola).

• BC represents line 'x = 1'.

• AD represents line 'x = 4'.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

➻ y² = x

➻ y = $\bf{\pm\:\sqrt{18}}$

Sɪɴᴄᴇ,

ABOCD is in 1st Quadrant,

➻ y = $\bf{\sqrt{18}}$

✔ We need to calculate the area of region ABEF.

$\red\checkmark\:\:\bf{Area\:of\:ABEF\:=\:\int\limits_{1}^{4}\:y\:.\:dx}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\int\limits_{1}^{4}\:\sqrt{x}\:.\:dx}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\int\limits_{1}^{4}\:{(x)}^{\frac{1}{2}}\:.\:dx}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\bigg[\:\dfrac{(x)^{\frac{1}{2}\:+\:1}}{\frac{1}{2}\:+\:1}\:\bigg]_1^4}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\bigg[\:\dfrac{(x)^{\frac{3}{2}}}{\frac{3}{2}}\:\bigg]_1^4}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\dfrac{2}{3}\:\Big[\:(x)^{\frac{3}{2}}\:\Big]_1^4}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\dfrac{2}{3}\:\Big\{\:(4)^{\frac{3}{2}}\:-\:(1)^{\frac{3}{2}}\:\Big\}}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\dfrac{2}{3}\:(8\:-\:1)}$

$:\implies\:\:\bf{Area\:of\:ABEF\:=\:\dfrac{2}{3}\times{7}}$

$:\implies\:\:\bf\green{Area\:of\:ABEF\:=\:\dfrac{14}{3}\:square\:units}$

$\Large\bf\blue{Therefore,}$

The area of the region bounded by the curve 'y² = x' is $\bf{\dfrac{14}{3}\:sq.\:units}$.