Please see where I stopped and continue the problem. No need to re-write what I already did. Please format answers just like the question and then solve the problem 100% - TY :) 1 Step 4 We also know that if F and G are antiderivatives of f and g, respectively, then F + G is an antiderivi 8/10 4/5 x 10 - 3/ 7 P 3/7 x 7 + 13 / 4 p 13 / 4 x 4 . Step 5 Finally, if F is an antiderivative of f and C is any constant, then F + C is also an anti-derivative of f. have that the most general antiderivative of f(x) = 8x9 - 3x6 + 13x3 is F ( x ) = 2 Tutorial Exercise Find the most general antiderivative of the function. f(x ) = (x+4)(5x - 9) Step 1 It is not true that if F and G are antiderivatives of f and g, respectively, then F . G is an antiderivative of f . g. Therefore, in o this product, obtaining 5 2 5 x2 + 11X - 36 36 Step 2 We know that if k is any constant and n # -1, then an antiderivative of kx" is k(7 4 1 x + 1). Therefore, using C as the constant of integration, the most general antiderivative of f(x) = (x + 4)(5x - 9) = 5x2 + 11x - 36 is F(x) = Submit Skip (you cannot come back). 3 Tutorial Exercise Find f. Step 2 f "(t) = 12/\\t, f(4) = 41, f'(4) = 13 Since 13 = f '(4) we have that F ' ( 4 ) = 24 4 0 8 4 ) / 2 + C . Step 1 Therefore, Recall that = t-1/2. 13 = 48 0 48 + C - 35 9 - 35 = C . Now, since f "(t) = t' f ' (t ) = 2 24Vt + C. Step 3 We have f '(t) = 24t1/2 - 35. Therefore, f(t) = 1 16(t 35t + D. 16+3/2 - 35t Step 4 Since 41 = f(4) we have that f(4) = 16 7)312 - 35 ) + D. Therefore, 41 = + D = D. Submit Skip (you cannot come back) 4 A particle is moving with the given data. Find the position of the particle. a(t) = t2 - 4t + 8, s(0) = 0, s(1) = 20 s(t) = X