Area bounded by lines y = 2 + x, y = 2 - x and x = 2 is:

Question

Area bounded by lines y = 2 + x, y = 2 - x and x = 2 is: (A) 16 (B) 8 (C) 4 (D) 3

Tardigrade Admin · Accepted Answer

Given y = 2 + x and y = 2- x By solving these, we get x = 0 Now, the required area = ∫20 y dx = ∫20 [(2 + x) - (2-x)]dx = ∫20 (0+2x)dx = 2 . (x2/2) |20 = 2 . (4/2) = 4

Q. Area bounded by lines y = 2 + x, y = 2 - x and x = 2 is:

16

8

4

3

Given y = 2 + x and y = 2- x
By solving these, we get x = 0
Now, the required area = $\int_{0}^{2} y d x$
$= \int_{0}^{2} [(2 + x) - (2 - x)] d x$
$= \int_{0}^{2} (0 + 2 x) d x = 2. \frac{x ^{2}}{2} ∣ ∣_{0}^{2} = 2. \frac{4}{2} = 4$

Q. Area bounded by lines y = 2 + x, y = 2 - x and x = 2 is:

16

8

4

3

Given y = 2 + x and y = 2- x By solving these, we get x = 0 Now, the required area = ∫02​ydx =∫02​[(2+x)−(2−x)]dx =∫02​(0+2x)dx=2.2x2​∣∣​02​=2.24​=4

Given y = 2 + x and y = 2- x
By solving these, we get x = 0
Now, the required area = $\int_{0}^{2} y d x$
$= \int_{0}^{2} [(2 + x) - (2 - x)] d x$
$= \int_{0}^{2} (0 + 2 x) d x = 2. \frac{x ^{2}}{2} ∣ ∣_{0}^{2} = 2. \frac{4}{2} = 4$