Calculus Assignment: Integral 4 Questions and Solutions for Practice

This document provides a comprehensive solution set for four integral calculus problems. The solutions cover a range of concepts, including determining intervals where a function is both decreasing and concave upward, calculating the time when a particle changes direction based on its acceleration, finding the maximum displacement of a function, and calculating the volume of a shape generated by revolving an area around the y-axis. Each solution includes detailed steps and explanations, making it a valuable resource for students studying calculus and looking to practice and understand integral concepts. The document is designed to help students solidify their understanding of integral calculus and improve their problem-solving skills. Desklib provides access to this and many more resources.

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,

V = π∫
c
d
[ f ( y ) ]2
dy
Here, f(y) is equation of the curve which is rotated w.r.t y axis expressed in terms of y.
[c,d] the y limits of rotation.
Curve equation: y = x^2 => x = sqrt(y).
Now, intersection of x = 2 and y = x^2 is calculated.
y = 2^2 = 4.
The curve starts from (0,0). Hence, lower y limit is 0.
Hence, the y limits are [0,2].
Hence, volume V = π∫
0
4
[ √( y) ]2
dy = π y2
2 ∨¿0
4 ¿ = 8π.