If you do graph it, you’ll see that y=x^2 looks like a big smiley face with its vertex (the point where the curve changes direction) at x=0, y=0. For this sample problem, you need to calculate the area (in square units) between the vertex (x=0) and an imaginary straight line drawn from x=4 along the x-axis up to the curve.
A = ∫a,b f(x) dx → A = ∫0,4 x^2 dx
A = (x^3)/3 |0,4
A = [(4^3)/3] - [(0^3)/3]
A = 64/3 - 0 A = 21 1/3 square units
A = ∫a,b f(x) dx A = ∫1,3 x^3 dx A = (x^4)/4 |1,3 A = (3^4)/4 - (1^4)/4 A = 81/4 - 1/4 = 80/4 A = 20 square units
A = ∫a,b f(x) dx A = ∫-2,2 x^3 dx A = (x^4)/4 |-2,2 A = (2^4)/4 - (-2^4)/4 A = 16/4 - 16/4 A = 0 square units This answer may seem wrong, especially if you graph y=x^3: how can it have zero area under the curve? Mathematically speaking, any area below the x-axis is a negative area; since the areas above and below the x-axis are equal and cancel each other out, the result is A = 0.
A = ∫a,b f(x) dx A = ∫-2,4 x^3 dx A = (x^4)/4 |-2,4 A = (4^4)/4 - (-2^4)/4 A = 256/4 - 16/4 = 240/4 A = 60 square units Like with the previous example (y=x^3 from x=-2 to x=2), remember that the area below the x-axis is actually a negative area. In this case, though, there is a greater area above the axis, leading to the final positive result.
A = ∫a,b f(x) dx A = ∫1,4 1/(x^2) dx = ∫1,4 x^-2 dx A = (x^-1)/-1 |1,4 = -1/x |1,4 A = -1/4 - (-1/1) = -1/4 + 1 A = 3/4 square units
A = ∫a,b f(x) dx A = ∫1,2 x^2-2x+8 dx A = (x^3)/3 - (2x^2)/2 + 8x |1,2 A = (x^3)/3 - x^2 + 8x |1,2 A = [(2^3)/3 - 2^2 + 8(2)] - [(1^3)/3 - 1^2 + 8(1)] A = [8/3 - 4 + 16] - [1/3 - 1 + 8] A = 8/3 - 4 + 16 - 1/3 + 1 - 8 A = 5 + 7/3 = 7 1/3 square units
A = ∫a,b f(x) dx A = ∫0,3 x^2-3x+5 A = (x^3)/3 - (3x^2)/2 + 5x |0,3 A = [(3^3)/3 - (3(3^2))/2 + 5(3)] - [(0^3)/3 - (3(0^2))/2 + 5(0)] A = 27/3 - 27/2 + 15 - 0 = 24 - 27/2 A = 10 1/2 square units
