How do you calculate the derivative of int(2t-1)^3 dt  from t=x^2 to t=x^7?

Feb 15, 2017

${\left(2 {x}^{7} - 1\right)}^{3} - {\left(2 {x}^{2} - 1\right)}^{3}$

Explanation:

Let $f \left(t\right) = {\left(2 t - 1\right)}^{3}$. Then, if $F \left(t\right)$ an antiderivative of $f \left(t\right)$, it holds that:

$F ' \left(t\right) = f \left(t\right)$.

We also know that $\int f \left(t\right) \mathrm{dt} = F \left(t\right) + c$. But also,

$\left(F \left(t\right) + c\right) ' = f \left(t\right)$ because $\left(c\right) ' = 0$ for any $c \in \mathbb{R}$.

Therefore,

$\left[\int f \left(t\right) \mathrm{dt}\right] ' = f \left(t\right)$,

so evaluating the derivative of that integral is exactly the same as evaluating $f \left(t\right)$ at the two given $t$ points, then taking their difference (as the question says to evaluate from ${x}^{2}$ to ${x}^{7}$)

Finally,

${\left\{\left[\int f \left(t\right) \mathrm{dt}\right] '\right\}}_{{x}^{2}}^{{x}^{7}} = f \left({x}^{7}\right) - f \left({x}^{2}\right)$

$= {\left(2 {x}^{7} - 1\right)}^{3} - {\left(2 {x}^{2} - 1\right)}^{3}$

Note that further simplification is applicable (but not required) since this is a difference of cubes.

Mar 6, 2017

$= x \left(7 {x}^{5} {\left(2 {x}^{7} - 1\right)}^{3} - 2 {\left(2 {x}^{2} - 1\right)}^{3}\right)$

Explanation:

$\frac{d}{\mathrm{dx}} \left({\int}_{{x}^{2}}^{{x}^{7}} {\left(2 t - 1\right)}^{3} \mathrm{dt}\right)$

We can use the Fundamental Theorem of Calculus , Part 1:

$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}\right) = f \left(x\right)$

But if it is:

$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt}\right)$ where $u = u \left(x\right)$, then we must also use the chain rule:

$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right) = \frac{d}{\mathrm{du}} \left({\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt}\right) \frac{\mathrm{du}}{\mathrm{dx}} = f \left(u \left(x\right)\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

To make this particular integral fit within the FTC, we need also do some interval manipulation:

$\frac{d}{\mathrm{dx}} \left({\int}_{{x}^{2}}^{{x}^{7}} {\left(2 t - 1\right)}^{3} \mathrm{dt}\right)$

Splitting out the integration intervals:

$= \frac{d}{\mathrm{dx}} \left({\int}_{{x}^{2}}^{0} {\left(2 t - 1\right)}^{3} \mathrm{dt} + {\int}_{0}^{{x}^{7}} {\left(2 t - 1\right)}^{3} \mathrm{dt}\right)$

Switching the first interval limits:

$= - \frac{d}{\mathrm{dx}} \left({\int}_{0}^{{x}^{2}} {\left(2 t - 1\right)}^{3} \mathrm{dt}\right) + \frac{d}{\mathrm{dx}} \left({\int}_{0}^{{x}^{7}} {\left(2 t - 1\right)}^{3} \mathrm{dt}\right)$

Applying the FTC:

$= - {\left(2 \left({x}^{2}\right) - 1\right)}^{3} \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + {\left(2 \left({x}^{7}\right) - 1\right)}^{3} \frac{d}{\mathrm{dx}} \left({x}^{7}\right)$

$= {\left(2 {x}^{7} - 1\right)}^{3} \cdot 7 {x}^{6} - {\left(2 {x}^{2} - 1\right)}^{3} \cdot 2 x$

$= x \left(7 {x}^{5} {\left(2 {x}^{7} - 1\right)}^{3} - 2 {\left(2 {x}^{2} - 1\right)}^{3}\right)$

Maybe that can be simplified a bit more.