# Ranges and the Fundamental Theorem of Calculus 1

I'm going over a chapter by chapter review for my calculus final and discovered this problem:

$$y=\int_{\sqrt{x}}^{x^3}\sqrt{t}\sin{t}\;\mathrm dt$$

They split it up so that it became:

$$-\int_1^{\sqrt{x}}\sqrt{t}\sin{t}\;\mathrm dt + \int_1^{x^3}\sqrt{t}\sin{t}\;\mathrm dt$$

Why 1? How was that determined?

• Does the problem want you to take the derivative? The idea is that you need an actual real number as the lower limit to apply the FTC. Other than that, 1 was completely random. Put any other positive number there and try taking the derivative. You'll find that you get the same answer.
– Matt
Dec 4 '10 at 22:58
• The use of 1 here is only for reasons of convenience, I think; note that in your first expression, if $x$ is 1, then $y$ is zero. Dec 5 '10 at 2:21

It doesn't have to be $1$; all it needs to be is in the domain of the function.

What is going on is that you have the Fundamental Theorem of Calculus, which tells you that if $f(x)$ is continuous on an interval containing $a$, then the function $F(x)$ defined on that interval by: $$F(x) = \int_a^x f(t)\,dt$$ is differentiable, and in fact $F'(x) = f(x)$ for all $x$ in the interval. But this requires (i) the lower limit to be constant; and (ii) the upper limit to be the variable with respect to which you are taking the derivative.

So the first question is: what does one do if you have the variable in the lower limit instead of the upper limit? Well, that only requires you to use the property of the integral that says $$\int_a^b f(t)\,dt = -\int_b^a f(t)\,dt,$$ and the fact that $(-G(x))' = -G'(x)$ for any $G$. So if $$F_2(x) = \int_x^a f(t)\,dt$$ then $$F_2'(x) = \frac{d}{dx}\int_x^a f(t)\,dt = \frac{d}{dx}\left(-\int_a^x f(t)\,dt\right) = -\frac{d}{dx}\int_a^x f(t)\,dt = -f(x),$$ with the last equality holding by the FTC.

Next, what if the upper limit is not $x$, but a function of $x$, say $$F_3(x) = \int_a^{g(x)} f(t)\,dt\ ?$$ Then we use the Chain Rule: $$\frac{dF_3}{dx} = \frac{dF_3}{dg}\frac{dg}{dx} = \left(\frac{d}{dg}\int_a^{g(x)}f(t)\,dt\right)\frac{dg}{dx} = f(g(x))g'(x),$$ with the last equality again by the FTC.

What if the lower limit is the function and the upper limit the constant? We combine the two "tricks" above to get the derivative.

And finally, what if both upper and lower limit are functions? Then we use the property of the integrals mentioned by Huy: $$\int_a^b f(t)\,dt = \int_a^cf(t)\,dt + \int_c^b f(t)\,dt$$ for any $c$ such that $f(x)$ is defined and continuous on an interval containing $a$, $c$, and $b$. In the case at hand, they pick $1$ simply because it's an easy point; you can use any point on $[0,\infty)$ (cannot be a negative point because you have $\sqrt{t}$ in the integrand). Pick your favorite $a$ on $[0,\infty)$, and you have $$F(X) = \int_{\sqrt{x}}^{x^2}f(t)\,dt = \int_{\sqrt{x}}^a f(t)\,dt + \int_a^{x^2}f(t)\,dt = -\int_a^{\sqrt{x}}f(t)\,dt + \int_a^{x^2}f(t)\,dt$$ and each of the integrals on the right is an integral we know how to do using the FTC. So if we let the first integral be some $G(X)$ and the second be $H(X)$, then $F(X)=G(X)+H(X)$, so $F'(X) = G'(X)+H'(X)$, and we can find $G'(X)$ and $H'(X)$ using the methods outlined above.

They did it with $a=1$, but you could do it with $a=\pi$, or $a=4$, or any nonnegative $a$.

• Thanks Arturo! That was very thorough and informative. Dec 5 '10 at 2:44

You can take any lower limit, it doesn't need to be 1.

$\int_b^c f(x) dx = \int_a^c f(x) dx - \int_a^b f(x) dx$

Just think of it as the area under the curve.