Answer:
[tex]\int_{0}^{2}sin(x^{2})dx \approx 1units^2[/tex]
Step-by-step explanation:
First of all, the graph of the function [tex]f(x)=sin(x^2)[/tex] is shown in the first figure below. We need to calculate the area under the curve which is in fact the definite integral. From calculus, we know that [tex]f(x)=sin(x^2)[/tex] is non integrable, that is, it doesn't have a primitive, so we must use calculator to evaluate [tex]\int_{0}^{2}sin(x^{2})dx[/tex]. To do so, calculator uses the Taylor Series, so:
[tex]sin(x^{2})=\sum_{n=-\infty}^{+\infty}\frac{(-1)^{n}}{(2n+1)!}x^{4n+2}$[/tex]
You an use a calculator or any program online, and the result will be:
[tex]\int_{0}^{2}sin(x^{2})dx=0.804units^2[/tex]
Since the problem asks for rounding the result to the nearest integer, then we have:
[tex]\boxed{\int_{0}^{2}sin(x^{2})dx \approx 1units^2}[/tex]
The area is the one in yellow in the second figure.
