What happens if we integrate, then differentiate?

Question

For a positive number $a$ , let $I(a) = \int_0^a \! \left(4 - 2^{x^2} \right) \, dx.$

Then $\dfrac{dI}{da}=0$ when $a$ equals

$\dfrac{1 + \sqrt{5}}{2}$ ,

$\sqrt{2}$ ,

$\dfrac{\sqrt{5} - 1}{2}$ ,

$1$ .

Suggestion

We are looking for the stationary point of $I$ . What does this mean geometrically?

As $a$ increases, for which points does $I(a)$ increase and for which points does it decrease? What does this tell us about where the stationary points have to lie?

Solution

For a positive number $a$ , let $I(a) = \int_0^a \! \left(4 - 2^{x^2} \right) \, dx.$

Then $\dfrac{dI}{da}=0$ when $a$ equals

$\frac{1 + \sqrt{5}}{2}$ ,

$\sqrt{2}$ ,

$\frac{\sqrt{5} - 1}{2}$ ,

$1$ .

Approach 1

We want to find a stationary point of $I(a)$ , though it may not be clear exactly how to differentiate an expression in this form. As $I(a)$ is the (signed) area of the curve above the $x$ -axis between $0$ and $a$ , we want to find a point in which the area is neither increasing nor decreasing.

Let $f(x) = 4 - 2^{x^2}$ , so that $I(a) = \int_0^a f(x) \, dx$ . Then if $f(a)>0$ , the area must increase if $a$ increases by a small amount. Similarly if $f(a)<0$ , the area must decrease. This is by considering that the new areas added will be entirely above and entirely below the $x$ -axis in each case respectively. This picture also tells us that the rate at which the area changes is smallest when $f(a)$ is smallest. Therefore, the stationary points must occur when $f(a)=0$ .

So $\frac{dI}{da}= 0$ when $4- 2^{a^2} = 0$ . Hence $a^2 = 2$ , which gives that $a = \sqrt{2}$ (as $a$ is positive).

Therefore the answer is (b).

Approach 2

We may recall that, in many ways, integration can be seen as the opposite of differentiation. Therefore we may initially guess that $\frac{dI}{da}= 4- 2^{a^2}$ . To show that this is true, suppose that $4-2^{x^2}$ integrates to the function $F(x)$ . That is, $F$ is the function satisfying $F'(x)=4-2^{x^2}$ . Then we have that

I(a) = \int_0^a \! \left(4 - 2^{x^2} \right) \, dx = \bigl[ F(x) \bigr]_0^a = F(a) - F(0)

and so

\frac{dI}{da}= F'(a) = 4- 2^{a^2}

as $F(0)$ is a constant. Hence the derivative is $0$ when $a=\sqrt{2}$ , similarly to above.

As before, the answer is (b).

In the two arguments above, you may not be completely convinced by some of the assertions that have been made. For example, in the first approach, we asserted that if $f(a)>0$ , then the area must increase if $a$ increases by a small amount. What if $f$ decreases very quickly? Would there always be some amount by which we can increase $a$ so that the area still increases? In the second approach, if we cannot write down an expression for the integral $F(x)$ explicitly, must it always exist? This result is known as the fundamental theorem of calculus and is explored when studying mathematics at university.