Question about an integral.. Why is $a = $2?

Question

I found the item in monbukagakusho 2013 math B exam.

Consider the function $$F(x) = \int_a^x f(t)\ dt = x^3 - 2x^2 - x - a$$ with $a \ne 0$. Find $a$.

I looked at the answer sheet and $a = 2$, but I ended up with weird equation (?)

$0 = a^3-2a^2-2a$

so what should I do to get $a = 2$?

Thank you

Maybe on the formula defining $F(x)$ there is $+a$ and not $-a$, otherwise this makes no sense. — Crostul, Nov 06 '14 at 17:25
Do I need more coffee: why doesn't $F(a) = \int_a^a f(x) \ dx = 0$ ... hence the equation the OP poses for $a$? — Simon S, Nov 06 '14 at 17:29
Look at my answer. Ant and Thomas's answers are based on differentiating and integrating the same thing. But in doing this, they made a sign mistake... — Crostul, Nov 06 '14 at 17:35

Ant · Answer 1 · 2014-11-06T17:30:03.920

2

Differentiating both sides of the equation, with respect to $x$, you get

$$f(x) = 3x^2 - 4x - 1$$

Now you want to integrate $f$ between $a$ and $x$; this yields

$$F(x) = \int_a^x 3t^2 - 4t - 1 \ dt = |t^3 - 2t^2 - t|_a^x = x^3 - 2x^2 - x - a^3 + 2a^2 + a$$

But you want $F(x) = x^3 - 2x^2 - x + a$; this implies $-a^3 + 2a^2 = 0$, and given that $a \neq 0$, the only solution is $a = 2$

edited Nov 06 '14 at 17:30

answered Nov 06 '14 at 17:22

Ant

21,098

No, $F(x) = x^3-2x^2-x-a$, not $x^3-2x^2-x+a$. – Crostul Nov 06 '14 at 17:40
@Crostul I'm assuming the correct sign is $+a$, otherwise $a=2$ is not a solution – Ant Nov 06 '14 at 17:41
So, you changed the x with a in the end right ? And the sign should be + a instead of - a to get a=2, now i see. Thanks – Blue Tomato Nov 06 '14 at 17:45
@BlueTomato You're welcome, though I did not changed the $x$ with $a$. The "correct" expression for $F(x)$ is the one I first derived, you have to find when the expression given in the book is equivalent to that one and this happens when $a=2$ (if there is a plus sign) :-) – Ant Nov 06 '14 at 17:47

Thomas · Answer 2 · 2014-11-06T17:50:07.003

1

If $$ F(x) = \int_a^x f(t)\ dt = x^3 - 2x^2 - x - a $$ then $F'(x) = f(x)$. So $f(x) = 3x^2 - 4x - 1$. Now then $$ \int_a^x 3t^2 - 4t - 1\; dt = t^3 - 2t^2 - t\bigg]_a^x = x^3 - 2x^2 - x - (a^3-2a^2 - a). $$ So all you have to do is to solve $$ a^3 - 2a^2 - a = a. $$ That is $$ a^3 - 2a^2 - 2a= 0 $$ As noted in above/and below comments, $a= 2$ is not a solution. But if you had $F(x) = x^3 -2x^2 -x \color{red}+ a$, then you would need to solve $a^3 -2a^2 = 0$ which indeed has solutions $0$ and $s$.

edited Nov 06 '14 at 17:50

answered Nov 06 '14 at 17:26

Thomas

43,555

Your third equation is wrong: you have to solve $a^3-2a^2 - a = a$. – Crostul Nov 06 '14 at 17:38
@Crostul: I changed the answer. Thanks. – Thomas Nov 06 '14 at 17:50

score 1 · Answer 3 · answered Nov 06 '14 at 17:31

Since for all $x$ $$\int_a^x \mbox{something } dt = x^3- 2x^2 -x -a$$ hold, in particular this holds for $x=a$. So you get

$$0=\int_a^a \mbox{something } dt = a^3- 2a^2 -a -a$$

So your equation works, but the solution is not $a=2$.

But if you had

$$F(x) = \int_a^x \mbox{something } dt = x^3- 2x^2 -x +a$$

then you get (substituting $x=a$) the equation

$$a^2(a-2)=0$$

so maybe there is some mistake on the exercise.

Question about an integral.. Why is $a = $2?

3 Answers3