areafunctionsgraphing-functionsintegration
My book say that integration of $x^{1/2} \sin x$ is not possible, why is it so? Which functions do not have an anti derivative?
Does it mean that they do not have any area under the curve?
But that's not true since the graph says different.
(Source: https://www.desmos.com/calculator)
When is a function not integrable?
The word "integral" is used in two completely different senses. The first, called definite integral, has a simple geometric (or physical) interpretation, the second, called indefinite integral, is accessible only to people having the notion of "derivative of a function of one variable" in their repertoire. It is true that in the one-dimensional case there is a connection between the two notions. This connection is called the fundamental theorem of calculus.
(a) The definite integral: Given some sort of "intensity" $f(x)$ at each point $x$ of some domain $B$ (an interval, a sphere, a cube in ${\mathbb R}^n$, etc.), where $f(x)$ varies with $x$, one can ask for the "total effect" an agent of this intensity could have. This total effect is the integral of $f$ over $B$ and is denoted by $$\int_B f(x){\rm d}(x)$$ (or similar). From the geometric intuition behind it this quantity is a limit of Riemann sums, viz. $$\int_B f(x){\rm d}(x)\ =\ \lim_{\ldots} \sum_k f(\xi_k)\ \mu(B_k)\ ,$$ where the $B_k$ form a disjoint partition of $B$ into very small subdomains and $\mu$ denotes the natural geometric measure (length, surface area, $n$-dimensional volume) in the situation at hand.
(b) The indefinite integral: Given a function $t\mapsto f(t)$ on some interval $I\subset{\mathbb R}$ one may ask: Is this function the derivative of some other function $F(\cdot)$? The answer is yes, and in fact there is an infinite set of such functions $F(\cdot)$, whereby any two of them differ by a constant on $I$. This set of functions is called the indefinite integral of $f$ on $I$ and is denoted by $$\int f(t)\ dt\ .$$
(c) The fundamental theorem of calculus: Given two points $a$, $b\in I$ the difference $F(b)-F(a)$ has the same value for all functions $F\in\int f(t)\ dt$ and may as well be denoted by $$\int_a^b f(t)\ dt\ .$$ Now comes the theorem (and this is the real wonder, not the fact that taking the derivative of the antiderivative of $f$ gives back $f$): When $a<b$ then $$\int_{[a,b]} f(t)\ {\rm d}t = \int_a^b f(t)\ dt\ .$$ Here on the left side we have a limit of Riemann sums, and on the right side a difference of $F$-values.
Well, well, well, this question actually can go really deep into the elementary ideas of integration, or, in general, finding the area under a curve.
In the ancient times, and especially during the period when mathematicians such as Eudoxus and Archimedes lived, the main way in which one could calculate the area between non-linear curves was the Method of Exhaustion. We will examine, in short, how would this work on calculating the area of a circle.
Let a circle $\mathcal{C}$ of radius $R$ and let two squares, $A_1A_2A_3A_4$ (red) and $B_1B_2B_3B_4$ (blue) such that $A_1A_2A_3A_4$ is inscribed in $\mathcal{C}$ and $B_1B_2B_3B_4$ is circumscribed to $\mathcal{C}$. Then, we have the following situation:
Now, the next step is to enclose the circle in some "tighter" canonical polygons. An easy construction is to construct the octagons that are induced from the two squares $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$, as it is depicted in the following shape:
Then one can move a step forward to create a deca-exagon (16 angles canonical polygon) etc. It is clear the the red polygon is always underestimating the area of the circle and the blue polygon is always oversetimating it.
Now, it is relatively easy, using the know formulas for the area of canonical polygons, to prove that both the red and the blue area do converge to the area of the circle. However, since back then calculus was not something "known" with the modern-day meaning (numbers were always treted as geometrical objects and not as abstract notions) they used the following form of Eudoxus' principle:
Eudoxus' Principle: Given two quantities $A,B$, if, from the first quantity $A$ we substract a quantity $C$ that is larger that the half of $A$ and, then, from the remnant ($A-C$ in modern notation) we substract a quantity that is larger than it half and we continue this way, then, after a finite number of steps, the remnant quantity will be less than $B$.
So, in that framework, it was sufficient to prove that each construction of a new $2^n-$sided polygon was "cutting" more than half of the error - which can be proved, but has a long and tiring proof and, hence, is out of the spectrum of this answer. After that, "convergence" (this is a modern term; it was not used back then) was ensured. What was missing was to show that these two areas (blue and red) were both "converging" to the area of the circle. To do so (since calculus and, hence, limits, were unknown) an argument - again tiring, in the general case - of double contradiction was used.
It is remarkable that, in the same way one could find the volume of the sphere and several other 3D shapes.
It is called Riemann Integral, however, in order to find the very solid geometric foundation of the notions of integral and integration we should investigate the so-called Darboux Upper and Lower Sums.
Let us consider a bounded function $f:[a,b]\to\mathbb{R}$ - boundedness is necessary for Rieman-integration; else, what would be the meaning of the area under that curve? Then, let a partition $\mathcal{P_n}=\{a=x_0<x_1<\dots<x_n=b\}$ of the interval $[a,b]$ and also let $m_k=\inf\limits_{x\in[x_k,x_{k+1}]}f(x)$ and $M_k=\sup_{x\in[x_k,x_{k+1}]}f(x)$, $k=0,1,\dots,n-1$. Then we define the lower, $L(f,\mathcal{P_n})$, and upper $U(f,\mathcal{P_n})$, Darboux sums for that partition and function as follows: $$\begin{align*} L(f,\mathcal{P}_n)&=\sum_{k=0}^{n-1}m_k(x_{k+1}-x_k)\\ U(f,\mathcal{P}_n)&=\sum_{k=0}^{n-1}M_k(x_{k+1}-x_k) \end{align*}$$
Also, we give the following definition:
Thinner partition: A partition $\mathcal{Q}$ is called thinner than $\mathcal{P}$ if: $$\mathcal{P}\subseteq \mathcal{Q}$$
Now, as we go forward to thinner partitions $\mathcal{P_k}$, with $k>n$, we say that $f$ is Riemann integrable if: $$\lim_{k\to\infty}U(f,\mathcal{P_k})=\lim_{k\to\infty}L(f,\mathcal{P_k})=I\in\mathbb{R}$$ for every sequence of "decreasing" partitions (every partition of the sequence is thinner than every previous one).
Note now that, in this way, Rieman integration is "just" a more generalised implementation of the Method of "Exhaustion" that ancient mathematicians used. Instead of canonical polygons, we do use the polygons induced from the partition of $f$'s domain we choose and these polygons are, in a way inscribed and circmuscribed to that curve. Also, thinner partitions are, in a sense, the same thing as going form a canonical $n-$polygon to another canonical $m-$polygon, with $m>n$. Moreover, what we seek, in order to be able to say that a function is Riemann integrable, is for the two limits of the sequences of the estimations given by the Darboux sums do converge to the same number, which was also the wanted in the Method of Exhaustion, in spite of not being proved with modern day calculus but with "old-fashioned" - however, interesting - geometrical arguments.
So, to conclude, a function is not Riemann integrable exactly when the Method of Exhaustion cannot be applied to the curve that represents it graph.
However, this does not apply to Lebesgue integrability... (I don't feel this was the case the OP requested, however, I can extend to Lebesgue integration, if requested).
In a "palimpseston" (Παλίμψηστον) - a script written on parchement that has been scratched and re-wrtiten in a direction vertical to the previous one - that was found in the early 20th century in an orthodox monastery close to Constantinople, after a small "adventurous" trip and, eventually, the proper scientific investigation, it was found that, the original script was a letter that Archimedes had writen to Eratosthenes in which, among other things, he was explaining the way though which he could find the results he should prove, however, he notes that he does not consider this method to be a method of prooving his results, just a way to "get a felling" of what the goal is.
In one of his examples he describes the way he found the area of a parabolic part - see this article in wikipedia: https://en.wikipedia.org/wiki/Archimedes_Palimpsest - and at some point he mentions that the reason he cannot consider this a proof is that of using a "summation of segments over an interval of real numbers" - this is in modern terms, not the ones used by Archimedes! So, with respect to the historical frame, Archimedes was very close to the notion of infinite, non-countable summation that we give to the Riemann integral, nowadays.
Best Answer
$$\int x^{1/2}\sin(x)dx=\int x^{1/2}\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!} dx=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+5/2}}{(2k+\frac52)(2k+1)!}.$$
The series converges for all $x$.
By substitution then by parts, you have
$$\int x^{1/2}\sin(x)dx=2\int t^2\sin(t^2)dt=-t\cos(t^2)+\int \cos(t^2)dt,$$ where the last integral is known as the Fresnel cosine function.
To get a numerical value of the definite integral, Simpson's method will perform better.