If f(x) =∫x^2 sin(t^2) dt from[0,x], find f′(x)

Answer 1

f(x)=∫x²sin(t²)dt from [0,x] can be rewritten as follows:

f(x)=x²(∫sin(t²)dt).

As "x" is not a variable in this integrand (we are integrating with respect to t), we can pull the x² out in front and treat it as a constant during the integration. We then employ the chain rule:

f'(x)=d/dx(x²)(∫sin(t²)dt +x² d/dx(∫sin(t²)dt)

(for the purpose of formatting, I have excluded writing the limits of integration, [0,x])

So, f'(x)=[2x(f(x)/x²]+x²sin(x²)

=2(f(x)/x)+x²sin(x²).

As you can see, we have provided f'(x) in terms of f(x). The reason being that ∫sin(t²)dt, though innocent looking enough, is incredibly difficult to solve without going far beyond the scope of this problem. It even has a special name: a Fresnal Integral.