Evaluate the following two integrals by interpreting each in terms of areas.
∫_0^3▒〖(𝑥−1)〗 𝑑𝑥
To interpret this integral in terms of areas, we can visualize the function f(x) = x-1 on the interval [0,3]. This function is a straight line with a y-intercept at -1 and a slope of 1.
The integral ∫_0^3▒(𝑥−1) 𝑑𝑥 represents the area under the curve f(x) = x-1 from x=0 to x=3.
To calculate this area, we can split it into a rectangle and a triangle.
The rectangle has a base of 3 units and a height of 0-1 = -1 units (since the function is below the x-axis in this interval). The area of the rectangle is base x height = 3 x -1 = -3.
The triangle has a base of 3 units and a height of 2 units (distance from the x-axis to the line at x=3). The area of the triangle is 1/2 x base x height = 1/2 x 3 x 2 = 3.
Therefore, the total area under the curve f(x) = x-1 from x=0 to x=3 is -3 + 3 = 0.
So, ∫_0^3▒(𝑥−1) 𝑑𝑥 = 0.