Question

using integration find the area enclosed by the curve y= x-x^2 and positive x-axis from x=0 ​

Answer 1

Given:

A curve y= x-x²

To Find:

Area enclose by given curve and x-axis from x=0

Solution:

First of all, we have to find the intersection of given curve and x-axis

We know that, at x-axis, y=0

So, on putting y=0 in the equation of curve, we will get the the two points where curve meet the x-axis

$\longrightarrow\rm{0= x-x^{2}}$

$\longrightarrow\rm{x-x^{2}=0}$

$\longrightarrow\rm{x^{2}-x=0}$

$\longrightarrow\rm{x(x-1)=0}$

$\longrightarrow\rm{x=0,1}$

So, given curve intersect x-axis at x=0 and x=1

Therefore, we have find the area enclosed by the curve y= x-x² and positive x-axis from x=0 to x=1

$\rule{190}{1}$

Now, on applying integration, we get

$\longrightarrow\rm{\displaystyle\int\limits^1_0 {y}\,dx}$

$\longrightarrow\rm{\displaystyle\int\limits^1_0 {(x-x^{2})}\,dx}$

$\longrightarrow\rm{\Bigg[\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\Bigg]^1_0}$

$\longrightarrow\rm{\Bigg(\dfrac{(1)^{2}}{2}-\dfrac{(1)^{3}}{3}\Bigg)-\Bigg(\dfrac{(0)^{2}}{2}-\dfrac{(0)^{3}}{3}\Bigg)}$

$\longrightarrow\rm{\Bigg(\dfrac{1}{2}-\dfrac{1}{3}\Bigg)-\Bigg(\dfrac{0}{2}-\dfrac{0}{3}\Bigg)}$

$\longrightarrow\rm{\Bigg(\dfrac{3-2}{6}\Bigg)-0}$

$\longrightarrow\rm{\dfrac{1}{6}\:sq\:units}$

Hence, the required area is 1/6 sq. units

Answer 2

AnsWer :

1/6 sq units.

To Find :

Area of enclosed by the curve y = x - x².

Solution :

We have,

• Enclosed curve y = x - x².
• With positive x - axis from x = 0.

A/Q,

• When, given curve pass through x - axis, it means should be y = 0.

$\tt : \implies y = x - {x}^{2} .$

$\tt : \implies 0 = x - {x}^{2} .$

$\tt : \implies 0= x(1 - x) .$

$\tt : \implies x = 0 \: \: \red{or} \: \: x = 1.$

Now, We have Value of x.

★ Let use Integration to find Enclosed area ( W.r.t.x )

$\tt : \implies \int\limits_{0}^{1}y \: dx .$

$\tt : \implies \int\limits_{0}^{1}(x - {x}^{2} ) \: dx .$

$\tt : \implies \int\limits_{0}^{1}x \: dx - \int\limits_{0}^{1} {x}^{2} \: dx .$

$\tt : \implies \Bigg( \dfrac{ {x}^{2} }{2} \Bigg) _{0}^{1} - \Bigg(\dfrac{ {x}^{3} }{3} \Bigg)_{0}^{1} .$

$\tt : \implies \Bigg(\dfrac{ {(1)}^{2} }{2} - \dfrac{ {(0)}^{2} }{2} \Bigg) - \Bigg( \dfrac{ {(1)}^{3} }{3 } - \dfrac{ {(0)}^{3} }{3} \Bigg)$

$\tt : \implies \Bigg[ \dfrac{1}{2} - \dfrac{1}{3} \Bigg]$

$\tt : \implies \Bigg[ \dfrac{3 - 2}{6} \Bigg]$

$\tt : \implies \dfrac{1}{6} \: sq \: units.$

Therefore, the required enclosed area is 1/6 sq units.