Integral with help of fundamental theorem of calculus

Question

When I solve problems like this

$$ \frac{\mathrm{d}}{\mathrm{d} x}\int_{x}^{x^2}\ln^3 t\,dt $$

I usually would set $$ F(x)=\int_{0}^{x^2}\ln^3 t\,dt $$

and then with help of the fundamental theorem of calculus I would get:

$$ \int_{x}^{x^2}\ln^3 t\,dt=F(x^2)-F(x) $$

and then apply chain rule:

$$ \frac{\mathrm{d} }{\mathrm{d} x}\int_{x}^{x^2}\ln^3 t\,dt= \cdots = \frac{6\ln^2 x^2}{x}-\frac{3\ln^2 x}{x} $$

but in this case I get discontinuity at x = 0 and I guess that makes my solution invalid. I dont know how to tackle this problem so I appreciate all help I can get.

Answer 1

Two points of confusion here:

• Unless the problem says differently, there's no reason for this to make sense at $x=0$. After all, $\ln x$ doesn't!
• $F(t)$ is not $\ln^3 t$. Rather, $F(t)$ is some function whose derivative is $F'(t) = \ln^3 t$. IOW, Andrew Li is correct: the expression simplifies to $2x\ln^3 x^2 - \ln^3 x$. (Which further simplifies to $(16x-1)\ln^3 x$.)

Answer 2

One part of the "Fundamental Theorem of Calculus" says that $\frac{d}{dx}\int_a^x f(t)dt= f(x)$ where $a$ can be any number.

To deal with $\frac{d}{dx}\int_x^{x^2} f(t)dt$, I would first break the integral into $\int_x^a f(t)dt+ \int_a^{x^2}dt= -\int_a^x f(t)dt+ \int)a^{x^2} f(t)t$ where, again, $a$ is any number.

Certainly, the "Fundamental Theorem of Calculus says that $\frac{d}{dx}\left(-\int_a^x f(t)dt\right)= -f(x)$. For the other integral, let $u= x^2$ so that can be written as $\int_a^u f(t)dt$. Then by the "fundamental Theorem of Calculus", together with the "chain rule" for differentiation, $$\frac{d}{dx}\int_a^u f(t)dt= \left(\frac{du}{dx}\right)\frac{d}{du}\int_a^u f(t)dt= \frac{du}{dx} f(u)= 2x f(x^2)$$

Putting those together, $\frac{d}{dx}\int_x^{x^2} f(t)dt$$= 2xf(x^2)- f(x)$. In this example, $f(t)= \ln^3(t)$ so that $$\frac{d}{dx}\int_x^{x^2} \ln^3(t)dt= 2x\ln^3(x^2)- \ln^3(x)$$.