Question

6) Find the area bounded by the curve y = x2 and the line y=x+6.​

Answer 1

Solution:

here, the given equations are

y = x^2  -----1

and y=x+6   -----2

To get the limits of the boundary we have to solve this equations.

Putting the value of y from equation 1 to equation 2 we get

x^2 = x +6

x^2 -x -6 =0

x^2 -3x +2x -6 =0

x(x -3)+2(x -3)=0

(x -3)(x +2) =0

x = 3 and -2

now, area bounded by the curve y = x2 and the line y=x+6 is

integration (x +6 -x^2)dx with limit -2 to 3

(x^2/2 +6x -x^3/3) with limit -2 to 3

[9/2 +18 - 27/3 -4/2 +12 +8/3]

[9/2 +8/3 -27/3+ 30- 2]

[9/2 -19/3+28]

(27-38 +168)/6

157/6 unit

Answer 2

EXPLANATION.

Area bounded by the curve,

⇒ y = x². - - - - - (1).

⇒ y = x + 6. - - - - - (2).

Put the value of equation (1) in equation (2), we get.

⇒ x² = x + 6.

⇒ x² - x - 6 = 0.

Factorizes the equation into middle term splits, we get.

⇒ x² - 3x + 2x - 6 = 0.

⇒ x(x - 3) + 2(x - 3) = 0.

⇒ (x + 2)(x - 3) = 0.

⇒ x = - 2  and  x = 3.

As we know that,

Concepts :

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

$\sf \displaystyle Area = \int\limits^b_a y \ dx = \int\limits^b_a f(x)dx$

Using this concepts in this question, we get.

$\sf \displaystyle Area = \int_{-2}^{3} [(x + 6) - x^{2} ] dx$

$\sf \displaystyle Area = \int_{-2}^{3} (x) dx + \int_{-2}^{3} 6 \ dx - \int_{-2}^{3} (x^{2} ) dx$

$\sf \displaystyle Area = \bigg[\frac{x^{2} }{2} + 6x - \frac{x^{3} }{3} \bigg]_{-2}^{3}$

In definite integration first we put upper limits then we put lower limits in the expression, we get.

$\sf \displaystyle Area = \bigg[\frac{(3)^{2} }{2} + 6(3) - \frac{(3)^{3} }{3} \bigg] - \bigg[ \frac{(-2)^{2} }{2} + 6(-2) - \frac{(-2)^{3} }{3} \bigg]$

$\sf \displaystyle Area = \bigg[\frac{9}{2} + 18 - 9 \bigg] - \bigg[2 - 12 - \frac{(-8)}{3} \bigg]$

$\sf \displaystyle Area = \bigg[\frac{9}{2} + 18 - 9 \bigg] - \bigg[2 - 12 + \frac{8}{3} \bigg]$

$\sf \displaystyle Area = \bigg[\frac{9}{2} + 9 \bigg] - \bigg[- 10 + \frac{8}{3} \bigg]$

$\sf \displaystyle Area = \frac{9}{2} + 9 + 10 - \frac{8}{3}$

$\sf \displaystyle Area = \frac{9}{2} - \frac{8}{3} + 19$

$\sf \displaystyle Area = \frac{27 - 16 + 114 }{6}$

$\sf \displaystyle Area = \frac{141 - 16}{6}$

$\sf \displaystyle \boxed{Area = \frac{125}{6} \ sq.units.}$