Q.112. Area using Integration.

Question 112.
Find the area of the region bounded by the lines
y + 3x = 4,     x + y = 4     and     x - y = 0
by taking integral over x or y as appropriate.

Answer 112.
Required area of the triangle

= sum of area of two triangles formed by the lines x - y = 0 and x + y = 4 and the y-axis less area of the triangle formed by the line y + 3x = 4 and the y-axis with limits as under.
[You shall understand the limits on drawing the figure]
= ∫(y=1 to 2) ydy + ∫(y=2 to 4)(4 - y)dy - ∫(y=1 to 4) (4 - y)/3 dy
= [y^2/2] (y=1 to 2] + [4y - y^2/2] (y=2 to 4) - (1/3)(4y - y^2/2] (y=1 to 4)
= (2 - 1/2) + (8 - 6) - (1/3)(8 - 7/2)
= 3/2 + 2 - 3/2
= 2
--------------------------------------…
Verification:
Sjifting origin to (1, 1), the vertices have new coordinates (-1, 3), (0, 0) and (1, 1)
Area of triangle with these vertices
(1/2) modulus of determinant
l-1 3l
l1 1l
= 2.

LINK to YA!