Integrations Unit Assessment

Question

If the area under a curve is given by x³/3 - x²/2 - 2x, find the exact area between this curve, the x-axis and its x-intercepts.

Answer 1

The area under the curve is given by, A(x) = x³/3 - x²/2 - 2x

Therefore, the equation of the curve is y = f(x) = dA/dx = x² - x - 2

x-intercepts: Setting y = 0, we get,

x² - x - 2 = 0

or, (x+1)(x-2) = 0

∴ x = -1, 2

The parabola y = x² - x - 2 opens upward, since the coefficient of x² is positive. So the enclosed area is below x-axis (negative).

Therefore, the exact area between this curve, the x-axis and its x-intercepts is

- ∫ f(x)dx, with lower limit -1 and upper limit 2

= - [A(2) - A(-1)]

= [A(-1) - A(2)]

= [(-1)³/3 - (-1)²/2 - 2(-1)] - [(2)³/3 - (2)²/2 - 2(2)]

= 9/2

Answer 2

x^3/3 -x^2/2 -2x = F(x)

take the derivative, then factor

F'(x)=f(x) = x^2 -x -2 = (x-2)(x+1) = 0

set factors = 0, solve for x

x =-1 and 2 are the x intercepts

evaluate F(x) from -1 to 2

complete the square to find the vertex

x^2 -x +1/4 -2 -1/4

= (x-1/2)^2 - 9/4 vertex = (1/2,-9/4) of an upward opening parabola with x intercepts -1 and 2

area all below the x axis = very very roughly bh/2= 3 x 9/4/2 = maybe about a little more than 27/8= 3 3/8 visually by geometry, graph f(x)= F'(x) and look at it

F(2)= 8/3 - 4/2 -4 = -3 1/3

F(-1) = -1/3 -1/2 +2 = 1 1/6.

Area = F(2)-F(-1) = -3 1/3 -1 1/6 = -4 1/2

the negative sign means the area is below the x axis

Area = 3 1/3 + 1 1/6

Area = exactly 4 1/2= 4.5 square units

if your exact integral calculation is in the general range of the estimate, it gives you more confidence you have the correct answer

Answer 3

Find the roots of this cubic polynomial. Factor out x, you'll be left with a quadratic. If you use the quadratic formula, you will see the roots of this polynomial are -1.812, 0, 3.312. You may verify the roots using desmos

Understand how the graph is laid out in desmos

You can then easily find the area under the graph by break into two intervals [-1.812,0] and [0,3.312]

Use the integration formula to find the area in each of the above two intervals to find the resulting net area