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(a). If $f$ is continuous on $[a,b]$ and $\int_a^x f(t) dt = 0$ for all $x \in [a,b]$, show that $f(x) = 0$ for all $x \in [a,b]$.

(b). If $f$ is continuous on $[a,b]$ and $\int_a^x f(t)dt = \int_x^b f(t)dt$ for all $x \in [a,b]$, show that $f(x)=0$ for all $x\in [a,b]$.

Work so far:

For (a), I think I am supposed to use Leibniz's rule and differentiate both sides and say $f(x)d/dx(x) - f(a)d/dx(a) = 0,$ so $f(x)-0=0$ and $f(x)=0.$ For (b) I think I am supposed to use Leibniz's Rule and differentiate both sides and get $f(x)d/dx(x) - f(a)d/dx(a) = f(b)d/dx(b) - f(x)d/dx(x)$, thus $f(x) - 0 = 0 - f(x)$, $2f(x) = 0$, and $f(x) = 0$ I going about this correctly?

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I've put the problem statement in LaTeX- please check I did not change the meaning of what you wrote. What does $f(x)\frac{d}{dx}$ mean? Do you want it to mean the derivative of $f(x)$ with respect to $x$? This is more commonly denoted $\frac{d}{dx}f(x)$ or $\frac{df(x)}{dx}$. – KReiser Aug 6 '12 at 18:38
Apart from notational issues, your calculations are correct. – André Nicolas Aug 6 '12 at 18:43
yes i meant the derivative of f(x) with respect to x – Rocket 12 Aug 6 '12 at 18:43

Yes, you've solved these accurately. As a small point, I wouldn't call these Leibniz' rule problems, as that relates to taking partial derivatives of multivariable functions in a direction orthogonal to the path of integration. This is an advanced calculus technique you may not have encountered yet.

Rather, this problem is really just using the fundamental theorem of calculus, which says for example that $\int_a^x f(t) dt=F(x)-F(a)$ with $F$ an antiderivative of $f$; then it's immediate from the rules of differential calculus that the derivative of the left-hand side with respect to $x$ is $f(x)$.

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Some hints:

a) Recall the first part of the Fundamental Theorem of Calculus $$f(x) = \frac{d}{dx}\int_a^x f(t)\ dt$$

b) Write $$\int_{a}^{x}f(t)\ dt + \int_{b}^{x}f(t)\ dt = 0$$ and do something similar to part a.

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