Question

Find the area bounded by the curve 2y=-x+8, x axis and the ordinate x=2 and x=4

Answer 1

EXPLANATION.

Area bounded by the curve,

⇒ 2y = -x + 8.

Ordinates at points x = 2 and x = 4.

As we know that,

Equation 2y = -x + 8.

⇒ 2y = 8 - x.

⇒ y = 8 - x/2.

$\sf Area = \int\limits^b_a {y} \, dx$

Put the value of y = 8 - x/2 in equation, we get.

$\sf Area = \int\limits^4_2 {\dfrac{8 - x}{2} } \, dx$

$\sf Area = \dfrac{1}{2} \int\limits^4_2 {(8 - x)} \, dx$

$\sf Area = \dfrac{1}{2} \int\limits^4_2 {8} \, dx - \dfrac{1}{2} \int\limits^4_2 {x} \, dx$

$\sf Area = \dfrac{1}{2} \bigg[ 8x - \dfrac{x^{2} }{2} \bigg]_2^4$

$\sf Area = \dfrac{1}{2} \bigg[8(4) - \dfrac{(4)^{2} }{2} \bigg] - \dfrac{1}{2} \bigg[ 8(2) - \dfrac{(2)^{2} }{2} \bigg]$

$\sf Area = \dfrac{1}{2} \bigg[ 32 - 8 - 16 + 2 \bigg]$

$\sf Area = \dfrac{1}{2} \bigg[ 34 - 24 \bigg]$

Area = 5 S.q units.

                                                                                               

MORE INFORMATION.

Symmetrical area.

If the curve is symmetrical about a coordinate axis ( or a line or origin ) then we find the area of one symmetrical portion and multiply it by the number of symmetrical portions to get the required area.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

Find the area bounded by the curve 2y = x+8, x axis and the lines x=2 and x=4.

$\large\underline\purple{\bold{Solution :- }}$

$\large\underline{\bold{❥︎Step :- 1 }}$

☆ Curve Sketching:-

Gɪᴠᴇɴ :

Linear equation,

$➢  \: \begin{gathered}\bf\blue{2y\: = \:x\: + \:8} \\ \end{gathered}$

❶ Substituting 'x = 0' in the given equation, we get

$\tt \ \: :  ⟼ 2y = 0 + 8$

$\tt \ \: :  ⟼ 2y = 8$

$\tt \ \: :  ⟼ y = 4$

❷ Substituting 'y = 0' in the given equation, we get

$\tt \ \: :  ⟼ 2 \times 0 = x + 8$

$\tt \ \: :  ⟼ 0 = x + 8$

$\tt \ \: :  ⟼ x = - \: 8$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf - 8 & \sf 0 \end{array}} \\ \end{gathered}$

$\large\underline{\bold{❥︎Step :- 2 }}$

$\tt \ \: :  ⟼ Required \: Area = \: \int_2^4 \: ydx$

$\tt \ \: :  ⟼ \int_2^4 \: \dfrac{x + 8}{2} \: dx$

$\tt \ \: :  ⟼ \dfrac{1}{2} \tt \ \: \int_2^4 \: (x + 8) \: dx$

$\tt \ \: :  ⟼ \dfrac{1}{2} \bigg[\dfrac{ {x}^{2} }{2} + 8x \bigg]_2^4$

$\tt \ \: :  ⟼ \dfrac{1}{2} \bigg((8 + 32) - (2 + 16) \bigg)$

$\tt\implies \:11 \: square \: units$