Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is a problem in our homework, I've never seen an integral written like that.. what does this question mean and how to solve it? Help please!! :(

Given that $\displaystyle 3x^2-12=\int_a^xf(t)~dt$, find a formula for $f(x)$ and a value of $a$.

Thanks a lot!

share|cite|improve this question
Hint: take the derivative of both sides of the equation with respect to $x$. – Hanul Jeon Feb 3 '13 at 4:18
but why the question gives the right side as f(t)dt and not f(x)dx? – user42624 Feb 3 '13 at 4:20
Because you can’t use the same symbol for the dummy variable of integration and for a limit of that integration; that would make no sense. Remember, for each value of $x$ the integral $\int_a^xf(t)dt$ is simply a number, and it would be the same number if you wrote $\int_a^xf(u)du$ instead. – Brian M. Scott Feb 3 '13 at 4:21
i got f(x)=6x, but what to do next? Like what does it help to find a? – user42624 Feb 3 '13 at 4:33
up vote 2 down vote accepted

HINT: This is a problem about the fundamental theorem of calculus. It says that if


then $F'(x)=f(x)$. You’ve been given $F(x)$; use the fundamental theorem to find the $f(x)$. To find $a$, notice that $\int_a^xf(t)~dt=0$ when $x=a$. For what $x$ is your $F(x)$ equal to $0$?

share|cite|improve this answer

As @tetori noted, Take the derivative of both sides. This is allowed becuase the conditions of Fundmental Theorem satisfied for it. So $$(3x^2-12)'=\left(\int_a^x f(t)dt\right)'=f(x)$$ So $$6x=f(x)$$ Now we have $$3x^2-12=\int_a^x 6tdt=3t^2\big|_a^x=3x^2-3a^2$$ So $3a^2=12$.

share|cite|improve this answer
Fundamental Theorem of Calculus, and Babak, to the rescue! +1 :-) – amWhy Feb 3 '13 at 13:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.