1
$\begingroup$

Question :

Let $$f:[a,b]\rightarrow \mathbb R$$ be continuous and $$g:[c,d]\rightarrow \mathbb R$$ be differentiable . Define $$\psi(x) := \int_a^{g(x)} f(t)dt $$ . Prove that $\psi$ is differentiable and compute the derivative .

My Attempt :

Define $$F(x)=\int_a^x f(t)dt$$ Then by Second Fundamental Theorem of Calculus , we have $F$ is differentiable and $F'=f$ at the points of continuity of $f$ . Then $$\psi(x)=(g\circ F)(x)$$

Both being differentiable , their composition is so and hence $\psi$ is differentiable. And by Chain Rule we have , $$\psi'(x)=g'(F(x))F'(x) \\ =(g'(F(x)))\cdot f(x)$$ Upto this is I think ok. But now how to simplify $g'(F(x))$ $?$

$\endgroup$
  • $\begingroup$ In your last line of the question what is $h$ ? $\endgroup$ – Empty Oct 3 '15 at 16:32
  • $\begingroup$ @S.Panja-1729 : Typo Sir,typo. $\endgroup$ – user118494 Oct 3 '15 at 16:43
3
$\begingroup$

Your function is $\psi(x)=F(g(x))-F(a)$ Where $F(x)$ is a primitive of $f(x)$ so the derivative is: $$ \psi'(x)=F'(g(x))g'(x)=f(g(x))g'(x) $$

$\endgroup$
3
$\begingroup$

You have a mistake. The composition is

$$\psi (x) = \left( {F \circ g} \right)(x)$$

and hence you have

$$\eqalign{ & \psi '(x) = F'(g(x))g'(x) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = g'(x)f(g(x)) \cr} $$

$\endgroup$
-1
$\begingroup$

You did the composition wrong! The integral is $F\circ g(x)$ and not $g\circ F(x)$.

$\endgroup$

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.