Use your calculator to evaluate the limit from x equals 0 to 2 of the sine of x squared, dx. Give your answer to the nearest integer.

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.