Need help catagorizing Calculus 1 problems.?

Question - 1:

i) You are given f '(x); and you need to find f(x) for the given conditions.

ii) Given f '(x) = 4{(x)^(1/2)} + 4{(x)^(3/2)}

iii) Now integrate this and you will get f(x) = Integral of f '(x) + C

iv) Use the given conditions, at x = 1, f(x) = 10; solve this you will get the value of C

Substitute this value of C in the f(x) equation in step (3) ==> You have your end answer.

Question - 2:

i) First integrate g"(t); Let this be A. You get g'(t) = A + c

ii) Using the given conditions, t = 4 and g'(4) = 7, evaluate for the value of c

iii) Plug this value of c, you get g'(t).

iv) Then, integrate g'(t) obtained in (iii) above. Let this be B; So you get g(t) = B + k

v) As previous, using the given conditions, g(4) = 20, evaluate for k

So you are done with end answer.

∫ √(x) * (4+4x) dx

∫ [ 4√(x) + x^(3/2) ] dx

4 * x^(3/2)/(3/2) + x^(5/2)/(5/2) + C

4 * (2/3) * x^(3/2) + x^(5/2) * (2/5) + C

f(x) = (8/3) * x^(3/2) + x^(5/2) * (2/5) + C ====> f(1)=10

10 = (8/3) * 1^(3/2) + 1^(5/2) * (2/5) + C

10 = (8/3) + (2/5) + C

10 = (40/15) + (6/15) + C

10 = (46/15) + C

10 - (46/15) = C ====> C = 104/15

f(x) = (8/3) * x^(3/2) + x^(5/2) * (2/5) + 104/15

-----------------------------------------------------

∫ 3 / [ √ t ] dt

∫ 3 * t^(-1/2) dt

3 * t^(1/2)/(1/2) + C

g ' = 6 * t^(1/2) + C ====> g'(4) = 7

7 = 6 * 4^(1/2) + C

7 = 6 * 2 + C

7 = 12 + C ===> C = -5

g ' = 6 * t^(1/2) - 5

∫ 6 * t^(1/2) - 5 dt

6 * t^(3/2)/(3/2) - 5t + C

4 * t^(3/2) - 5t + C ====> g(4) = 20

20 = 4 * 4^(3/2) - 5*4 + C

20 = 4 * 8 - 20 + C

20 = 32 - 20 + C

20 +20 -32 = C

C = 8

g(t) = 4 * t^(3/2) - 5t + 8

arctan(u), arcsin(u), arccos(u) and so on. all turn a trignometric ratio into an perspective value arctan(a million) as an occasion is 40 5 tiers or (pi/4) arcsin(.5) is 30 tiers a) ought to be superb for all arctan values when you consider that they bypass from minus to plus infinity