# How do you use the second fundamental theorem of Calculus to find the derivative of given int sqrt(x^3-2x+6) dx from [-2, x^2]?

Mar 27, 2016

Now letting $x = t$ the given integral becomes ${\int}_{2}^{{x}^{2}} \sqrt{{t}^{3} - 2 t + 6} \mathrm{dt}$
Accordingly $\frac{d}{\mathrm{dx}} {\int}_{-} {2}^{{x}^{2}} \sqrt{{t}^{3} - 2 t + 6} \mathrm{dt} = 2 x \sqrt{{x}^{6} - 2 {x}^{2} + 6}$ [Put t= ${x}^{2}$ then $\frac{\mathrm{dt}}{\mathrm{dx}}$ = 2x