The following is a problem from Ch. 14, "The Fundamental Theorem of Calculus", from Spivak's Calculus
- Find a function $g$ such that
(ii) $\int_0^{x^2} tg(t)dt=x+x^2$
(Notice that $g$ is not assumed continuous at $0$.)
Here is my solution, which is broadly the same as the solution manual solution, but I am hung up on certain details.
Let $$f(x)=\int_0^{x^2} tg(t)dt$$ $$h(x)=xg(x)$$ $$f_1(x)=\int_0^x h$$ $$f_2(x)=x^2$$
Then $f=f_1 \circ f_2$ and $f'(x)=f_1'(f_2(x))f_2'(x)$.
Given some $x$, if $h$ is continuous at $x$ then we can apply the first fundamental theorem of calculus (FTC1) to obtain
$$f'(x)=x^2g(x^2)\cdot 2x=2x^3g(x^2)$$
Since $f(x)=x+x^2$ then we have
$$f'(x)=2x^3g(x^2)=1+2x$$
$$g(x^2)=\frac{1}{2x^3}+\frac{1}{x^2}$$
$$g(x)=\frac{1}{2x\sqrt{x}}+\frac{1}{x}$$
Note that $g$ is not defined at $0$, so there no way to write $xg(x)$ for $x=0$. We can however, defined $g$ as
$$g(x)=\begin{cases} \frac{1}{2x\sqrt{x}}+\frac{1}{x}, \text{ if } x\neq 0 \\ 0, \text{ if } x=0 \end{cases}$$
Then
$$h(x)=\begin{cases} \frac{1}{2\sqrt{x}}+1, \text{ if } x \neq 0 \\ 0, \text{ if } x=0 \end{cases}$$
I believe $h$ looks something like
Is $h$ integrable? It is not bounded near $0$. In Spivak the definition of integrability is
A function $f$ which is bounded on $[a,b]$ is integrable on $[a,b]$ if $\sup\{L(f,P)\}=\inf\{U(f,P)\}=\int_a^b f$.
Note that if $l(x)=\sqrt{x}+x$ then $l'(x)=h(x)$. Thus, since $h$ is continuous on $(0,\infty)$ we can apply FTC2 and have
$$\int\limits_{x_0}^{x^2} h= l(x^2)-l(x_0)=x+x^2-\sqrt{x_0}-x_0$$
The issue is, what justifies allowing $x_0$ to be $0$, as the solution manual allows?
