Question

find the area of the region bounded by the curve y=9x and y=3x​

Answer 1

We have, y  = 9x and y = 3x

Y  = 3 (3x) = 3y to solve

⇒ y = 0 or 3

when y = 0, x = 0 and when y = 3, x = 1 

Points are (0,0) and (1,3) 

Parabola y  = 9x and line y = are shown in the graph

आकृति ज्ञात करें, छायांकित क्षेत्र का क्षेत्रफल।   

Answer 2

Question:

Find the area of the region bounded by the curve y² = 9x and the line y = 3x

Answer:

Area bounded = 1/2 square units

Step-by-step explanation:

Given:

Curve y² = 9x and the line y = 3x

To Find:

The area bounded by the curve

Solution:

Here the given curve is a parabola,

y² = 9x having vertex (0,0) and given line

y = 3x-----(1)

Substitute the value of y from from equation 1,

(3x)² = 9x

9x² = 9x

x² = x

x² - x = 0

x (x - 1) = 0

Hence,

x = 0, x = 1

When x = 0, y = 0 and when x = 1, y = 3

Hence the points of intersection are (0,0) and (1,3)

Now area bounded the curve and line is given by,

$\sf Required\:area=\int\limits^1_0 {\sqrt{9x}-3x } \, dx$

(∵ y² = 9x, y = √9x)

Finding the value,

$\sf \implies \int\limits^1_0 {3\sqrt{x}-3x } \, dx$

$\sf \implies 3\times \int\limits^1_0 {\sqrt{x} } \, dx-3\times \int\limits^1_0 {x} \, dx$

$\sf \implies 3\times \int\limits^1_0 {x^{1/2} } \, dx-3\times \int\limits^1_0 {x} \, dx$

We know,

$\sf \int\limits {x^{n} } \, dx =\dfrac{x^{n+1}}{n+1}$

Therefore,

$\sf \implies 3\times \bigg[\dfrac{x^{1/2+1}}{1/2 + 1}\bigg]^1_0 -3\times \bigg[\dfrac{x^{1+1}}{1+1}\bigg]^1_0$

$\sf \implies 3\times \bigg[\dfrac{x^{3/2}}{3/2}\bigg]^1_0 -3\times \bigg[\dfrac{x^{2}}{2}\bigg]^1_0$

$\sf \implies 3\times \bigg[\dfrac{2}{3}\times 1-0\bigg] -3\times \bigg[\dfrac{1}{2}\bigg]$

$\sf \implies 2-\dfrac{3}{2}$

$\sf \implies \dfrac{1}{2}\:units^{2}$

Hence the area bounded by the curve and the line is 1/2 square units.