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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!

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Hint: take the derivative of both sides of the equation with respect to $x$. –  tetori 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

2 Answers 2

up vote 2 down vote accepted

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

$$F(x)=\int_a^xf(t)~dt\;,$$

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 $0$?

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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 $3a^2=12$.

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

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