I'm looking for a fun (not too many tedious calculations) calculus one problem that uses the concept that, after subsitution, you have two integrals of diffent functions with different limits, but equal area. For example:
Those two regions are the same. I'm thinking of a problem that asks the student to prove two regions are of equal area, and once they set up the integrals they can see it is a case of subsitution.
Do you know of such a problem?
On
In general there are two ways to compute $E(X^2)$ where $X$ is a random variable with density $f(x)$; for simplicity I'll say $X$ always takes values between $0$ and $a$, so $f(x) = 0$ when $x < 0$ or $x > a$. One is to simply take $\int_0^a x^2 f(x) \: dx$. The other is to find the density $g$ of the random variable $Y = X^2$; this is $g(y) = f(\sqrt{y})/(2\sqrt{y})$. Then
$$ E(Y) = \int_0^{a^2} y g(y) \: dy = \int_0^{a^2} {\sqrt{y} \over 2} f(\sqrt{y}) \: dy. $$
These two integrals are related by the change of variables $y = x^2$. You can generate many examples from this. For example take $f(x) = 6x(1-x)$ on $0 \le x \le 1$; then this gives
$$ \int_0^1 6x^3 - 6x^4 \: dx = \int_0^1 3y - 3y^{3/2} \: dy. $$
Although the inspiration for this comes from probability (which I'm teaching currently, which probably explains why this came to mind) the result is true even if $f$ takes on negative values or doesn't integrate to one.
On
The following is part of the standard machinery for proving the basic properties of logarithms, if one chooses to introduce them by using an integral.
For any number $c \ge 1$, define $L(c)$ by saying that $L(c)$ is the area under the curve $y=1/x$, above the $x$-axis, from $x=1$ to $x=c$.
Then $L(ab)$ is the area from $1$ to $ab$. This is the area from $1$ to $a$, plus the area from $a$ to $ab$.
But the area from $a$ to $ab$ is $$\int_a^{ab}\frac{dx}{x}$$ Make the substitution $x=ua$. We find that $$\int_a^{ab}\frac{dx}{x}=\int_1^b \frac{du}{u}$$
Conclusion: $L(ab)=L(a)+L(b)$.
Edit: Here is another example, more usable. Let $a$ and $b$ be positive. Then the area of the top half of the ellipse $x^2/a^2+y^2/b^2=1$ is the same as the area of the top half of the circle with radius $\sqrt{ab}$.
On
If we make the substitution $t=\frac{u}{1+u}$ in the beta function defined by the following first integral, we get the second:
$$B(p,q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}\;\mathrm{d}t=\int_{0}^{\infty }\frac{% u^{p-1}}{(1+u)^{p+q}}\;\mathrm{d}u.$$
Since the relation $B(p,q)=\frac{\Gamma (p)\Gamma (q)}{\Gamma (p+q)}$, for $p=q=\frac{1}{2}$ yields
$$B(1/2,1/2)=\frac{\left( \Gamma (1/2)\right) ^{2}}{\Gamma (1)% }=\pi,$$
we obtain
$$\int_{0}^{1}\frac{1}{\sqrt{t\left( 1-t\right) }}\;\mathrm{d}% t=\int_{0}^{\infty }\frac{1}{\left( 1+u\right) \sqrt{u}}\;\mathrm{d}u=\pi.$$
How about calculating the area, $S$, of one quarter of the unit circle?
First note that $S = \int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x} $. The substitution $x=\sin(u)$ (hence, ${\rm d}x = \cos(u) \,{\rm d}u$) gives $$ \int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x} = \int_0^{\pi /2} {\sqrt {1 - \sin ^2 (u)} \cos (u)\,{\rm d}u} = \int_0^{\pi /2} {\cos ^2 (u)\,{\rm d}u}. $$ The integral on the right can be computed as follows. $$ \int_0^{\pi /2} {\cos ^2 (u)\,{\rm d}u} = \int_0^{\pi /2} {\frac{{1 + \cos (2u)}}{2}\,{\rm d}u} = \frac{\pi }{4} + \frac{1}{2}\int_0^{\pi /2} {\cos (2u)\,{\rm d}u} = \frac{\pi }{4}, $$ where the last equality follows from $$ \int_0^{\pi /2} {\cos (2u)\,{\rm d}u} = \frac{1}{2}\int_0^\pi {\cos (t)\,{\rm d}t} = 0. $$ Thus we have shown that $S=\pi/4$, which is a non-trivial result.
EDIT: As another example, using the the substitution $y=e^{-x}$, we have $$ \int_0^\infty {e^{ - x} \,{\rm d}x} = \int_1^0 {y\frac{{\,{\rm d}y}}{{ - y}}} = \int_0^1 {1\,{\rm d}y} = 1. $$