Question

5x^3+40= the integral of f(t) dt from a to x solve for a

Answer 1

$5x^3+40=\displaystyle\int_a^xf(t)\,\mathrm dt$

Differentiate both sides with respect to $x$:

$\dfrac{\mathrm d}{\mathrm dx}\left[5x^3+40\right]=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_a^xf(t)\,\mathrm dt$
$15x^2=f(x)$

This means

$\displaystyle\int_a^x15t^2\,\mathrm dt=5x^3+40$
$5t^3\bigg|_{t=a}^{t=x}=5x^3+40$
$5(x^3-a^3)=5x^3+40$
$-5a^3=40$
$a^3=-8$
$a=-2$