Find each of the definite integrals given the function f(@) shown below. (Note that the slope of the function changes at c = -4) -5 - 2 -1 0 Choose... So f(x) dx Choose.. Choose... + -24 J's f(x) dx 6 Choose... + -15 l f(x) dx Choose... + 52 28 J of(x) dx Choose... # -4 24 So f(x) dx Choose.. + -10 -14 Choose... + 14 10Problem 1 (10 points). Consider the following problem: Approximate 8.993 using differ entials. A student submitted the solution below. It contains several mistakes. Identify all and explain how to x them. Then give the correct solution to the problem. Let f($) = 3:3 so that we are approximating f(8.99). 'We know that 8.99 is close to 9, and the differential is d3: = 0.01. 'We have HQ) = 729. Therefore Ag; = dy = f(9)d:r = 72900.01] = 7.29. Since f'($) = 3332, we also have f'(9) = 243. So we have 8.993 = f(8.99) = 243 x 7.29 = 1771.47. Problem 2 (4 points). Explain in your own words why we talk about an antiderivative of a function instead of the antiderivative. 4. Problem 1 (4 points). Find f (2 + x? 4 (a: 2?) d3: using geometry. 0 Hint: Split the integral rst and then identify which two geometric shapes the areas under the functions have. n 3 What is f(a) given that f" (x) = cos(x) and f'(7) = 1 and f( 7) = 3? ed O a. f(x) = - cos(a) out of O b. f(x) = - cos(ac ) + 2 + 2 - T question O c. f(ac) = - cos(a) + 2 + 4- T O d. f(x) = -cos(a) + ac O e. f(x) = cos(a) + x + 4 - 1We will find h(t) given that h"(t) = 3t + cos(t) and h' (0) = 3 and h(0) = -1. First, the indefinite integral of h"(t) is (3/2)t^2+sin(t)+c . . Clone 2 + Ch Since h' (0) = 3 and also h'(0) = 3+C C we have C = COS(1) +D 1 + C , (1/2)t^3+3t-cos(t)+D Next, the indefinite integral of h' (t) is -cosit)+D D (1/2)t^3+31+cos(1) +D Since h(0) = -1 and also h(0) = D+1 D- 3 we have D D-1 Therefore, h(t) = =+3 + 3t - cos(t)