Question

area bounded by parabola y=4x^2,x=0 and y=1,y=4 is

Answer 1

We need to find area bounded by the parabola,

$\longrightarrow y=4x^2$

or,

$\longrightarrow x=\pm\dfrac{\sqrt y}{2}$

Since the area is bounded by $x=0$ too, we need to check,

$\longrightarrow\dfrac{\sqrt y}{2}>0$

$\Longrightarrow y>0$

and,

$\longrightarrow-\dfrac{\sqrt y}{2}>0$

which is impossible. [Recall $\sqrt y$ is positive square root of $y.$]

Hence the area bounded will be given by,

$\displaystyle\longrightarrow A=\int\limits_1^4\dfrac{\sqrt y}{2}\ dy$

$\displaystyle\longrightarrow A=\dfrac{1}{2}\cdot\dfrac{2}{3}\left[y^{\frac{3}{2}}\right]_1^4$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{7}{3}}}$

Answer 2

☆Given Question :-

• The area bounded by parabola y = 4x^2, x = 0 and y = 1, y = 4 is

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$\huge \orange{AηsωeR}✍$

☆Given :-

$\sf \: Parabola \: y = {4x}^{2} , x = 0, y = 1, y = 4$

☆To Find :-

• Area bounded between these curves.

☆Solution :-

⟼ y = 4x² represents a upper parabola symmetric along y - axis and having vertex at (0, 0).

⟼ y = 1 is a line parallel to x - axis intersecting the parabola y = 4x² at the points

$\sf \: ⟼(\dfrac{1}{2} ,1 ) \: and \: ( - \dfrac{1}{2} , 1)$

⟼ y = 4 is a line parallel to x - axis intersecting the parabola y = 4x² at the points

$\sf \: ⟼(1 , 4) \: and \: ( - 1 ,4 )$

⟼ Consider a horizontal strip of length = x and width dy in the first quadrant.

⟼ Area of this rectangular strip = x dy.

⟼ This rectangular strips moves from y = 1 to y = 4.

$\bf \:As \: y = 4x²$

$\bf \:  ⟼ {x}^{2} = \dfrac{y}{4}$

$\bf \:  ⟼ x = \dfrac{ \sqrt{y} }{2}$

$\bf \:Therefore, \: area \: of \: shaded \: region \: is$

$\bf \:Area = \sf\int\limits_{1}^{4}xdy$

$\bf\implies \:Area = \sf\int\limits_{1}^{4}\dfrac{ \sqrt{y} }{2} dy$

$\bf\implies \:Area = \dfrac{1}{2} \sf\int\limits_{1}^{4} {y}^{\frac{1}{2} } dy$

$\bf\implies \:Area =\dfrac{1}{2}[\dfrac{ {y}^{ \frac{1}{2} + 1} }{ \frac{1}{2} + 1} ] \ _{1}^{4}$

$\bf\implies \:Area =\dfrac{1}{2}[\dfrac{ {y}^{ \frac{3}{2}} }{ \frac{3}{2}} ] \ _{1}^{4}$

$\bf\implies \:Area = \dfrac{1}{3} [ {(4)}^{ \frac{3}{2} } - {(1)}^{ \frac{3}{2} } ]$

$\bf\implies \:Area = \dfrac{1}{3} [ {(2)}^{2 \times \frac{3}{2} } - 1 ]$

$\bf\implies \:Area = \dfrac{1}{3} [ {2}^{3} - 1 ]$

$\bf\implies \:Area = \dfrac{1}{3} [8 - 1 ]$

$\bf\implies \:Area = \dfrac{7}{3} \: square \: units$

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