Hi I need help on what is the answer to this problems? Below is the question and other concerns will be in the comment section. I hope you can help me and surely I will give you full positive feedback.

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1. Let f(x) = 710 - 4p - 12r. If F(x) satisfies F(x) = f(x) and F(0) = 2, what is F(x)? a. F(x) =x - x- 6r+0 c. F(x) = - r - 61+2 b. F(x) = x - r - 6x] +1 d. F(x) =x - x -6r] 2. What is the particular antiderivative of f(x) = a passing through the point (2, 3)? + 1 b. 1= 1 d. y= -T3. Evaluate and determine its particular antiderivative whose r-intercept is i. 7 = arcsin + C. y = arcsin + 2. b. y = arcsin d. y = arcsin 4. Evaluate / 6r(4x] - 7)dr. (4 - 7)+0 C. -(4x] - 7)+ 0 b. =(4x] - 7) + 0 d. (4x3 - 7) + C 5. Determine the general antiderivative of () = 3-1 I - 1 C. tarts+o b. d. a-r'+:+C 6. Which of the following integrals can be solved easily using u-substitution? a. x'sin (r? )dr c. x'sin (Ar] )dr b. ( x'sin (r) )dr d. x' sin (3r] )dr 7. Determine the general antiderivative of /(x) = a. In x]- 4 + C c. 3lor-4+0 b. = In |23- 4 + 0 d. = In|3-4+C 8. Evaluate r lar a. =(Inz) + C C. = Ins + C b. In (Inz) + C d. In (Inz) + C 9. What is / tan (Ar + 3)der? a. 4 In | sec (4x + 3)| + c c. In | sec (4r + 3)| + C b. = In | sec (4x +3)| +C d. = In | sec (4r + 3)|+ C10. Evaluate / (2r) esc (x3 - 5)dr. a. In Jesc (x] - 5) - cot (x3 - 5)| + C b. 2r? In | esc (x2 - 5) - cot (13 - 5)| + C c. x? In | esc (13 - 5) - cot (13 - 5)|+ C d. 2 In | ese (23 - 5) - cot (23 - 5)| +C Adr 11. Evaluate T? + 61 + 34 a. - arctan 5 =arctan (= + 3) +C b. 4 arctan d. 5 arctan 12. What is a. - arcsin +0 C. - arcsin 2 b. 3 arcsin + C d. 6 arcsin 13. Evaluate 5353 d.x. In 5 +0 b, In 3 +0 C. In 3 . +0 d. 3 In 5 +0 14. Determine the antiderivatives of f(x) = 3weeks sec (5x) tan (5x). 350c (5x) a. 5(3(5)) 350 (5x) 15 0 (5x) 5 In 3 +0 c. In 3 In 3 d, -+0 In 3 15. Evaluate era+10x-7(r + 5)dr. a. ex+0-7 +C b. er+-7 + C c. 20'+10-7 + C d. (2e) 3+10-740