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