Integral 4 Question Practice

Integral 4 Question Practice
1. If f(x) = 2x^3-3x^2-12x+18, when is it both decreasing and concave up?
Ans: The function f(x) is increasing in the interval where f’(x) > 0 and it is
concave upward when f’’(x) >0. Now, first the points where f’(x) = 0 and
f’’(x) = 0 are found.
f’(x) = 0 => 6x^2 – 6x – 12 = 0 => x^2 – x -2 = 0 => (x-2)(x+1) = 0 => x
= 2 and x = -1.
Taking a point in between -1 and 2
f’(1) = 6 – 6 -12 = -12 (<0).
Hence, f(x) is decreasing in (-1,2)
Now, f’’(x) = 0 => 12x – 6= 0=> x = ½.
Now, f’’(3) = 12*3 – 6 = 30 (>0).
f’’(-3) = -72 – 6 = -78 (<0).
Hence, f(x) is concave upward in the interval (1/2,∞)
Hence, f(x) is concave upward and decreasing in (1/2,2).
2. If the acceleration of an object is a(t)=4t-12 and at time 0 the velocity is
10, when is the particle changing direction?
Ans: velocity of the particle v(t) = ∫ ( 4 t−12 ) dt = 2t^2 -12t + c
Given, v(0) = 10 => 10 = c
Hence, v(t) = 2t^2 – 12t +10.
Now, velocity change occurs from the point when acceleration is 0.
Hence, 4t-12 = 0 => t = 3.
Hence, particle changing its direction at t = 3.
3. If there is an equation t^4-4t^3, what is the maximum displacement
between t=-2 and t=4? (NOTE: Check endpoints for answers)
Ans: displacement equation f(t) = t^4 – 4t^3.
Now, maximum displacement occurs when f’(t) = 0 and f’’(t) <0.
f’(t) = 0 => 4t^3 – 12t^2 = 0 => t-3 = 0 => t =3 and t =0.
f’’(t) = 12t^2 – 24t => f’’(3) = 36 (>0). (displacement is minimum).
f’’(0) = 0 (not-conclusive evidence).
Now, checking the endpoints.
f(-2) = 16 + 32 = 48.
f(4) = 4^4 – 4*4^3 = 0.
Hence, the maximum displacement occurs at t = -2 of the displacement
function.
4. What is the volume of a shape that consists of the area between y=x^2
and x=2 revolved around the y-axis?
Ans: The volume for revolution around y axis is given by,