Consider the expression y(x)=y0+∫axf(t)dt that arises from a first–order differential equation dxdy=f(x) with y(a)=y0. Which of the following best explains why this expression represents a unique curve passing through the point (a,y0)?
a) The Fundamental Theorem of Calculus states that all antiderivatives differ by a constant, so different values of that constant produce the same curve.
b) Because f(t) is continuous near a, the existence–uniqueness theorem for differential equations guarantees exactly one solution curve through (a,y0).
c) The definite integral fixes the constant of integration, so y(a)=y0 is satisfied without introducing any additional unknowns.
d) The mean value theorem guarantees that f(t) attains its average value on [a,x], forcing the graph through (a,y0).