# Fundamental theorem of Calculus

1. Mar 17, 2014

Suppose $a$ is a constant.

If $G(x) = \int_a^x \Big [ f(t) \int_t^x g(u) du \Big ] dt$, what is $G\,'(x)$?

My attempt,
$$G\,'(x) = f(x) \int_x^x g(u) du = 0,$$

and I am sure this is wrong.

Last edited: Mar 18, 2014
2. Mar 18, 2014

### Ray Vickson

$$G(x) = \int_a^x F(x,t) \, dt$$
in general, what would be your answer for $G'(x)$? What happens if you specialize this to
$$F(x,t) = f(t) \int_t^x g(u) \, du ?$$

3. Mar 20, 2014

Thank you for the hint. This is my attempt at the proof which I mimic from that of the single variable case. Please check the rigour in the proof.

Let
\begin{equation*}
G(x) = \int_a^x F(x,t) dt.
\end{equation*}

Now
\begin{align*}
G(x+h) - G(x)
&= \int_a^{x+h} F(x+h,t) dt - \int_a^x F(x,t) dt \\[3mm]
&= \int_a^x F(x+h,t) dt + \int_x^{x+h} F(x+h,t) dt - \int_a^x F(x,t) dt \\[3mm]
&= \int_a^x F(x+h,t) dt - \int_a^x F(x,t) dt + \int_x^{x+h} F(x+h,t) dt \\[3mm]
&= \int_a^x \big [ F(x+h,t) - F(x,t) \big ] \; dt + h F(x+h,c) \\[3mm]
\end{align*}

for some $c \in (x, x+h)$ by the mean value theorem for integrals. Therefore
\begin{equation*}
\frac{G(x+h) - G(x)}{h}
= \int_a^x \Big [ \frac{F(x+h,t) - F(x,t)}{h} \Big ] \; dt + F(x+h,c) \\[3mm]
\end{equation*}
and
\begin{align*}
G'(x) &= \lim_{h \to 0} \int_a^x \Big [ \frac{F(x+h,t) - F(x,t)}{h} \Big ] \; dt
+ \lim_{h \to 0} F(x+h,c) \\[3mm]
&= \int_a^x \frac{\partial}{\partial x} F(x,t) dt + F(x,x)
\end{align*}
since as $h \to 0, x + h \to x$ and $c \to x$.

Now if $F(x,t) = f(t) \int_t^x g(u) du$. Then

\begin{equation*}
\frac{\partial}{\partial x} F(x,t) = f(t) g(x)
\end{equation*}
and
\begin{equation*}
G'(x) = \int_a^x f(t) g(x) dt + f(t) \int_x^x g(u) du = \int_a^x f(t) g(x) dt.
\end{equation*}

4. Mar 20, 2014

### Ray Vickson

That is correct. There is a general formula for such problems that is often found in good calculus textbooks. Assuming everything is differentiable, we have:
$$\frac{d}{dx} \int_{a(x)}^{b(x)} F(x,t) \, dt = b'(x) F(x,b(x)) - a'(x) F(x,a(x)) + \int_{a(x)}^{b(x)} F_x(x,t) \, dt$$

The proof is essentially the same as what you gave.

5. Mar 20, 2014

I would like to go back a bit and look at:

If $\quad {\displaystyle G(x) = \int_a^{\psi(x)} f(t) dt } \quad$ then $\quad G\,' (x) = \psi\,' (x) f(\psi(x)).$

My attempt at the proof, which again mimic the proof of the basic case:

Proof

\begin{align*}
G(x+h) - G(x)
&= \int_a^{\psi(x+h)} f(t) dt - \int_a^{\psi(x)} f(t) dt \\[3mm]
&= \int_{\psi(x)}^{\psi(x+h)} f(t) dt \\[3mm]
&= \big [\psi(x+h) - \psi(x) \big ] f(\psi(c))
\end{align*}
for some $c \in (x, x+h)$ by the mean value theorem.

Therefore
\begin{equation*}
\frac{G(x+h) - G(x)}{h}
= \frac{\psi(x+h) - \psi(x)}{h} f(\psi(c))
\end{equation*}
and
\begin{align*}
G\,'(x) &= \lim_{h\to 0} \frac{\psi(x+h) - \psi(x)}{h} f(\psi(c)) \\[3mm]
&= \psi'(x) f(\psi(x))
\end{align*}
since as $h \to 0, x + h \to x$ and $c \to x$.

The bit that I am not comfortable with is going from

${\displaystyle \int_{\psi(x)}^{\psi(x+h)} f(t)} dt \quad$ to $\quad {\displaystyle \big [\psi(x+h) - \psi(x) \big ] f(\psi(c)) }$

Perhaps more explanation is needed here to make it rigorous?

Thank you sir for stating the general case. I saw that in a textbook a long time ago but I did not jot it down and I have been trying to find it ever since. All the Calculus textbooks that I refer to do not state the result.

6. Mar 20, 2014

### Dick

The formula Ray is talking about is quoted here http://en.wikipedia.org/wiki/Leibniz_integral_rule. And for justifying your step, you gave the reason. It's a formulation of the mean value theorem. You may need to state some assumptions about continuity, but other than that it's perfectly rigorous.