Can someone help me? Need to find the area (Integration/Calculus)?

The answer is as follows:

Alternative method to find area:
Area enclosed
= [Area formed by (0,0), (1,3) and (4,0)] - [Area formed by (0,0, (2,2) and (4,0)]
= (1/2) × 4 × 3 - (1/2) × 4 × 2
= 6 - 4
= 2 (sq. units)

https://gyazo.com/b903a339116c8aba9e982d72a675287c
Region enclosed by the three curves is (0,0) , point (1) and point (2)
Split the triangle into two areas.

(1) is the intersection of y=3x and y=-x+4
3x=-x+4
4x=4
x=1
y=3x = 3(1) = 3

(1) is (1,3)

(2) is the intersection of y=x and y=-x+4
x=-x+4
2x=4
x=2

(2) is (2,2)

Area bounded by y=3x and y=x from x=0 to 1:
= & [0,1] (3x-x) dx
= & 2x
= 2 x^2 /2
= x^2
F(x) = x^2
F(1) = 1
F(0) = 0
F(1)-F(0) = 1 --------(1) first area

Area bounded by y=-4+x and y=x from 1 to 2
= & [1,2] (-x+4-x) dx
= & [1,2] (-2x+4) dx
= -2x^2 /2 + 4x

F(x) = -x^2+4x
F(2) = 4
F(1) = 3
F(2)-F(1) = 1 ---------(2) second areaa

Area bounded by the three curves is (1)+(2)
= 1+1
= 2

Hint:
Sketch graphs
Find the points of intersections
Set up integrals

　
Region:
https://www.desmos.com/calculator/mnrzq5whp5

For 0 ≤ x ≤ 1 ---> x ≤ y ≤ 3x
For 1 ≤ x ≤ 2 ---> x ≤ y ≤ −x+4

Area = ∫ [0 to 1] (3x−x) dx + ∫ [1 to 2] (−x+4−x) dx
Area = ∫ [0 to 1] (2x) dx + ∫ [1 to 2] (4−2x) dx
Area = x² | [0 to 1] + (4x−x²) | [1 to 2]
Area = (1 − 0) + ((8−4)−(4−1))
Area = 2

——————————————————————————————

Alternate method:

For 0 ≤ y ≤ 2 ---> y/3 ≤ x ≤ y
For 2 ≤ y ≤ 3 ---> y/3 ≤ x ≤ −y+4

Area = ∫ [0 to 2] (y−y/3) dy + ∫ [2 to 3] (−y+4−y/3) dy
Area = ∫ [0 to 2] (2y/3) dy + ∫ [2 to 3] (4−4y/3) dy
Area = (y²/3) | [0 to 2] + (4y−2y²/3) | [2 to 3]
Area = (4/3 − 0) + ((12−6)−(8−8/3))
Area = 2