9. Find the area of the region bounded by the parabola y = x² and y = |x|

4 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Given that equation of

curve $y = x^{2}$

line $y = | x | = {x, i f x \geq 0 & - x, i f x ∠ 0}$

Since the line passes through A&B in Ist and IInd quadrants

the equation must satisfy

$\Rightarrow y = x^{2}$

$\Rightarrow x = x^{2}$ for Ist quadrant and

$\Rightarrow - x = x^{2}$ for IInd t quadrant

So, $\Rightarrow x^{2} - x = 0$ and $x^{2} + x = 0$

$\Rightarrow x (x - 1) = 0$ and $x (x + 1) = 0$

$\Rightarrow x = 0, 1$ $x = 0, - 1$

$x = 0, y = 0$

$x = 1, y = 1$ i.e, A has coordinate (1,1)

$x = - 1, y = 1$ i.e, B has coordinate (1,1)

Now, area of AODA = area (AOM)-area (ADOM)

$= \int_{0}^{1} y_{l i n e} d x - \int_{0}^{1} y_{c u r v e} d x$

$= \int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x = {[\frac{x^{2}}{2}]}_{0}^{1} - {[\frac{a^{3}}{3}]}_{0}^{1}$

$= \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} u n i t s^{2}$

$∴$ The required area of the region bounded by curve $y = x^{2}$ and line $y = | x |$ is $\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} u n i t s^{2}$

0 Upvotes 0 Downvotes

Similar Questions for you

Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

Raj Pandey

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) Þ c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

Raj Pandey

Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$