This document contains four practice questions on integrals. The questions cover topics such as decreasing and concave up functions, changing direction of a particle, maximum displacement, and volume of a shape. Each question is explained step by step with the corresponding answer.
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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)= 0and 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) =∫(4t−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 maximumdisplacement 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 revolvedaround the y-axis? Ans: The volume for revolution around y axis is given by,
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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=πy2 2∨¿0 4¿= 8π.