Evaluate the integral by interpreting it in terms of areas. Lower limit =0 Upper limit = 9 |10x-9|dx?

Put y = 9 |10x - 9|

The minimum of y = 0, at this point :
9 |10x - 9| = 0
10x - 9 = 0
x = 9/10

When x ≥ 9/10, y = 9(10x - 9)
Hence, when x = 9, y = 9[10(9) - 9] = 729

When x < 9/10, y= -9(10x - 9)
Hence, when x = 0, y = =9(0 - 9) = 81

The graph is shown in the following link :
https://c2.staticflickr.com/8/7321/27590722852_92691903f9_o.png

₉
∫ 9 |10x - 9|
⁰
= Area of ΔAOB + Area of ΔCDB
= (1/2) × OA × OB + (1/2) × CD × BD
= (1/2) × 81 × (9/10) + (1/2) × 729 × [9 - (9/10)]
= 2988.9

First graph
y = |10x - 9|
on the interval 0 ≤ x ≤ 9.

You should see that the area above the x-axis and below the graph forms two triangles. One very narrow one over the interval
0 ≤ x ≤ 9/10
and one very large one over the interval
9/10 ≤ x ≤ 10.

Use the geometric formula for the area of triangles to find the area of each of the triangles.

Then add those two areas to get find the integral.