<p>5x^3+40= antiderivative of a to x f(t)dt. Solve for value of a.</p>
<p>a.-2
b.2
c.1
d.-1
e.0</p>
<p>I'd say A, that's assuming that F(x) = 5x^3+40, where F(x) is the antiderivative of f(t) at x since the integral from a to x of f(t)dt = F(x) - F(A) by the fundamental theorem of calculus. But I don't see how the antiderivative of a function of t, can be related to the original function. I don't know, I'd say A though.</p>