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

Feb 15, 2016

See the explanation section, below.

Explanation:

If we are told to use the second part, then we are being asked to find the function first, then differentiate it. (If we were using the first part, we would not find the function at all.)

$g \left(x\right) = {\int}_{-} {2}^{{x}^{2}} \left({t}^{3} - 2 t + 6\right) \mathrm{dt}$

$= {\left[{t}^{4} / 4 - {t}^{2} + 6 t\right]}_{-} {2}^{{x}^{2}}$

$= \left[{\left({x}^{2}\right)}^{4} / 4 - {\left({x}^{2}\right)}^{2} + 6 \left({x}^{2}\right)\right] - \left[{\left(- 2\right)}^{4} / 4 - {\left(- 2\right)}^{2} + 6 \left(- 2\right)\right]$

$= {x}^{8} / 4 - {x}^{4} + 6 {x}^{2} + 12$

And, so

$g ' \left(x\right) = 2 {x}^{7} - 4 {x}^{3} + 12 x$

If we had used the first part, we would combine it with the chain rule to get:

$g ' \left(x\right) = \left({\left({x}^{2}\right)}^{3} - 2 \left({x}^{2}\right) + 6\right) \cdot \left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right]$

$= \left({x}^{6} - 2 {x}^{2} + 6\right) \left(2 x\right)$

$= 2 {x}^{7} - 4 {x}^{3} + 12 x$