# Past exam question on integral of an even function!

I am going through a past exam question which is a function $$f:\mathbb{R} \rightarrow \mathbb{R}$$ is defined by $$\begin{equation*} f(x) := \int_{1}^{x^2} e^{t^2} \; dt. \end{equation*}$$
1. Determine whether $$f$$ is even, odd or neither.
What I did was $$f$$ is an even function. Let $$h(t) = e^{t^2}$$ such that we have $$\int_{1}^{x^2} h(t) \; dt$$. Thus, $$\begin{equation*} \begin{split} f(-x) &= \int_{1}^{(-x)^2} h(t) \; dt \\ &= \int_{1}^{x^2} h(t) \; dt \\ &= f(x). \end{split} \end{equation*}$$
2. If $$f$$ bounded below?
I know for $$f$$ to be bounded below there is an $$m\in \mathbb{R}$$ with $$f(x)\geq m$$, $$\forall x\in \mathbb{R}$$. So clearly it is because $$f$$ is always positive so is bounded below by $$0$$.
3. Explain why $$f$$ is differentiable and calculate its derivative $$f'$$.

To find $$f'$$ we just differentiate both sides to get $$f'(x) = e^{(x^2)^2} = e^{x^4}$$ by the fundamental theorem of calculus, but how do I explain why it is differentiable?

• Hint: $(-x)^2$ is not equal to $-x^2$... Aug 8 '19 at 23:30
• Any function which depends on $x$ only through $x^{2}$ is an even function. Aug 8 '19 at 23:31
• As for part 2: if $x^2<1$ then the integral is negative, even though the integrand is positive. Aug 8 '19 at 23:32
• Under $u=-t$ integral from $1$ to $x^{2}$ becomes intergral from $-x^{2}$ to $-1$. Aug 8 '19 at 23:34
• I always thought since $f$ is even then the integral of it would be odd. Or am I off the right path there! Aug 8 '19 at 23:34

As many of the comments have indicated, a function is even if $$f(x) = f(-x)$$. This is easy to see as $$$$f(-x) = \int_1^{(-x)^2} \exp(t^2) dt = \int_1^{x^2} \exp(t^2) dt = f(x).$$$$
The function $$f$$ is bounded below, but it is not always positive. It is actually bounded below by $$\frac{-\sqrt{\pi}}{2}erfi(1)$$, which is the value of $$f(0)$$.
Last, the integral of an infinitely differentiable function $$h(t) = e^{t^2}$$ will also be differentiable. Using the fundamental theorem of calculus, sometimes called the Leibniz rule, the derivative is, \begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} \int_1^{x^2} \exp(t^2) dt\\ &= \exp((x^2)^2) \cdot \frac{d}{dx} (x^2)\\ &= 2x e^{x^4}. \end{align}
If you need a real proof of why it is differentiable (past simply finding its derivative), then you could for example use the definition of begin differentiable, and try to show that $$$$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$$$ is well defined for all $$x$$. In other words, you could compute the limit for general $$x$$, get the derivative written above, and that is sufficient proof of differentiability.
• Great thank you. Just one question how did you get $f(0) = -\frac{\sqrt{\pi}}{2}$ as the lower bound? Aug 9 '19 at 9:59
• Good catch, I was working too fast, the lower bound is slightly lower than that, it is $-\sqrt{\pi} \frac{erfi(1)}{2}$, I've fixed this in the answer. The integral is only negative when the $x \in (-1,1)$. Intuitively, it is zero at $x = \pm 1$, and $x^2$ is symmetric on the interval, so half way in between at $x = 0$ a max/min occurs. So that is one way to "know" where to look. Otherwise, you could use calculus to find the extrema points by seeing where the first derivative of the integral is zero. Aug 13 '19 at 7:20