Determine the area between the given curve, the x-axis and vertical indicated: y = x^3, x = -3 and x = 3?

Integral from -3 to 0 of x^3dx + Integral from 0 to 3 of x^3 dx = 2x^4/4 = 81/2

that area corresponds to the integral of x³ between x = -3 and x = 3.

The simple answer is that because the plain old cube function is odd and the interval is centered around zero, the function will have dot symmetry at the origin. This means that the area to the left of the y-axis and the area to the right will compensate each other and the integral will be 0 (areas that are below the x-axis are considered negative).

However, if it comes to actually calculating, it you would take the primitive (in this case, F(x) = x⁴/4) and calculate the variation between -3 and 3:

F(3) - F(-3) = (3)⁴/4 - (-3)⁴/4 = 0

x^3 + x^2 - 90x = 0, x(x^2 + x - ninety) = 0, x(x^2 + 10x - 9x - ninety) = 0, x(x(x + 10) - 9(x + 10)) = 0, x(x + 10)(x - 9) = 0. So y crosses the x-axis at x ? {-10,0,9}. Then our section is |?(x^3 + x^2 - 90x)dx [-10,0]| + |?(x^3 + x^2 - 90x)dx [0,9]| = |(x^4)/4 + (x^3)/3 - 45x^2 [-10,0]| + |(x^4)/4 + (x^3)/3 - 45x^2 [0,9]| = |10000/4 - 1000/3 - 4500| + |6561/4 + 729/3 - 3645| = |(7500 - 1000 - 13500)/3| + |(6561 + 972 - 14580)/4| = |-7000/3| + |-7047/4| = 7000/3 + 7047/4 = (28000 + 21141)/12 = 49141/12. -x^2 - 13x - 40 = 0, x^2 + 13x + 40 = 0, x^2 + 5x + 8x + 40 = 0, x(x + 5) + 8(x + 5) = 0, (x + 5)(x + 8) = 0. So y crosses the x-axis at x ? {-8,-5}. Then our section is |?(-x^2 - 13x - 40)dx [-8,-5]| = |-(x^3)/3 - 13(x^2)/2 - 40x [-8,-5]| = |(100 and twenty 5/3 - 325/2 + 2 hundred) - (512/3 - 416 + 320)| = |(250 - 975 + 1200)/6 - (512 - 1248 + 960)/3| = |475/6 - 224/3| = |(475 - 448)/6| = 27/6 = 9/2.