3
$\begingroup$

Let $f:[a, b]\rightarrow \mathbb{R}$ be continuous and let $F:[a, b]\rightarrow \mathbb{R}$ be defined by $F(x)=\int_a^x f(t) \,dt$. Then $F$ is differentiable whose derivative is $f$.

Now define $G$ on $[a, b]$ by $G(x)=\int_x^b f(t)\, dt$.

It can be shown, following the proof of above theorem, that $G$ is differentiable with derivative $-f$ (is this correct)? However, I wanted to see, how the assertion about $G$ follows from that of $F$ without following the proof of assertion for $F$? Any hint?

$\endgroup$
  • 1
    $\begingroup$ Use $-G = \intop_b^x f(x) dx$ $\endgroup$ – Shai Deshe Jan 19 at 14:51
3
$\begingroup$

HINT

Note that $$ A = \int_a^b f(t) dt $$ is a constant and $$ G(x) = \int_x^b f(t) dt = \int_a^b f(t) dt - \int_a^x f(t) dt = A - F(x) $$

$\endgroup$
4
$\begingroup$

Here's a hint: start with the fact that if $a \le x \le b$ then $$\int_a^b f(t) \, dt = \int_a^x f(t) \, dt + \int_x^b f(t) \, dt $$ which is a simple consequence of the definition of definite integrals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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