When are absolute values needed in the antiderivative? I just tackled a STEP I past paper question

Brenden Tran

Brenden Tran

Answered question

2022-06-20

When are absolute values needed in the antiderivative?
I just tackled a STEP I past paper question (Q4(iii), 2004) which asked me to determine the antiderivative for this integral:
1 ( x + 2 ) x 2 + 4 x 5 d x
which (after substituting x + 2 3 = sec ( t )) gives:
1 3 sec 1 ( x + 2 3 ) + C
This is also the answer given by Cambridge in the answer booklet. However, I also put this integral into an online integral calculator and one of the possible antiderivatives was
1 3 arcsin ( 3 x + 2 ) + C 1
This is (according to Desmos) is equivalent to:
1 3 sec 1 ( x + 2 3 ) + C 2
which is not at all the same function as the one I got. However, looking at the graphs (of the original functions and the antiderivatives), it would seem that the integral calculator is correct. I am presuming that Cambridge simplified the answer for STEP.
I can see that the quadratic in the radical has a minimum point at x = 2, but we haven't really been taught much about absolute values and discontinuities, except for
1 x d x = ln x + C,
which we were told was because ln(x) can't take negative values. Yet, arcsin(x) above can, and there's an absolute value there.
Could someone explain to me how all of this works, please?

Answer & Explanation

Eleanor Luna

Eleanor Luna

Beginner2022-06-21Added 19 answers

Step 1
You can learn a lot by really understanding the single antiderivative 1 t d t ..
We are tempted to interpret this as (1) 1 x 1 t d t = ln x , ( x > 0 )
but we know that the choice of the base point 1 is arbitrary. We could find another antiderivative by looking at 4 x 1 t d t = ln x ln 4.
Step 2
This ambiguity leads to the "up to an additive constant C", as both of these are acceptable antiderivatives when x > 0.
But what about when x < 0? Then 1 x 1 t d t doesn't converge - the domain of integration includes the point 0, near which 1/t shoots off to infinity. On the other hand, if x < 0, then (2) x 1 1 t d t = 1 x 1 t d t = ln ( x ) . ( x < 0 )
(The equality comes from substituting t t and then writing a b = b a .) Of course, we know this is really only true up to a constant C.
We know that (1) applies for all x > 0 and (2) applies for all x < 0. There is no antiderivative corresponding to x = 0, since every integral of the form
A 0 1 t d t
diverges. Thus a complete story would be to say that the antiderivative of 1 t is
1 t d t = { ln x + C if  x > 0 , ln ( x ) + C if  x < 0.
It just happens to be that this is given by ln | x | + C that's sort of a coincidence.
Jeffery Clements

Jeffery Clements

Beginner2022-06-22Added 3 answers

Step 1
Your antiderivative is wrong when x < 1 (its derivative has the wrong sign).
If you use the substitution ( x + 2 ) / 3 = sec ( t ) (I don't think you really used x = sec ( t )), your integral becomes
tan ( t ) 3 sec 2 ( t ) 1 d t
Now sec 2 ( t ) 1 = tan 2 ( t ), so it may be tempting to write this as d t / 3 = t / 3 + C leading to your answer. But this is wrong: if tan ( t ) < 0 (which corresponds to x < 5) we will have sec 2 ( t ) 1 = tan ( t ). So a correct answer would be
{ 1 3 sec 1 ( x + 2 3 ) + C  if  x > 1 1 3 sec 1 ( x + 2 3 ) + C  if  x < 5
which (with different C in the second case, but there's no need for the C's to be the same) could be written as
1 3 sec 1 ( | x + 2 | 3 ) + C
Of course, if for some reason you're only interested in x > 5 your answer would be OK.

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