# What is the derivative of  int_x^(x^2) t^3 \ dt  wrt x?

May 26, 2017

$\frac{d}{\mathrm{dx}} {\int}_{x}^{{x}^{2}} {t}^{3} \mathrm{dt} = {x}^{6} - {x}^{3}$

#### Explanation:

According to the ${2}^{n d}$ Fundamental Theorem of Calculus,
$\frac{d}{\mathrm{dx}} {\int}_{a}^{b} f \left(t\right) \mathrm{dt} = f \left(b\right) - f \left(a\right)$.

Using your example, $\frac{d}{\mathrm{dx}} {\int}_{x}^{{x}^{2}} {t}^{3} \mathrm{dt} = {\left({x}^{2}\right)}^{3} - {x}^{3} = {x}^{6} - {x}^{3}$.

May 26, 2017

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} = 2 {x}^{7} - {x}^{3}$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

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

(ie the derivative of an integral gives us the original function back).

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt}$

Initially we can manipulate the integral as follows (although we have chosen $0$ as the lower limit we could in fact choose any constant:

${\int}_{0}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} = {\int}_{0}^{x} {t}^{3} \setminus \mathrm{dt} + {\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt}$

And so:

${\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} = {\int}_{0}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} - {\int}_{0}^{x} {t}^{3} \setminus \mathrm{dt}$

And therefore differentiating we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} = \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} - \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{x} {t}^{3} \setminus \mathrm{dt}$

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

$u = {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

The substituting into the first integral we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} = \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{u} {t}^{3} \setminus \mathrm{dt} - \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{x} {t}^{3} \setminus \mathrm{dt}$

$\text{ } = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{du}} \setminus {\int}_{0}^{u} {t}^{3} \setminus \mathrm{dt} - \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{x} {t}^{3} \setminus \mathrm{dt}$

$\text{ } = 2 x \frac{d}{\mathrm{du}} \setminus {\int}_{0}^{u} {t}^{3} \setminus \mathrm{dt} - \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{x} {t}^{3} \setminus \mathrm{dt}$

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{2}} {t}^{3} \setminus \mathrm{dt} = 2 x \setminus {u}^{3} - {x}^{3}$
$\text{ } = 2 x \setminus {\left({x}^{2}\right)}^{3} - {x}^{3}$
$\text{ } = 2 x \setminus {x}^{6} - {x}^{3}$
$\text{ } = 2 {x}^{7} - {x}^{3}$

May 27, 2017

Another way to do this is to apply Leibniz's integral rule:

$\frac{d}{\mathrm{dx}} \left[{\int}_{a \left(x\right)}^{b \left(x\right)} f \left(x , t\right) \mathrm{dt}\right] = {\int}_{a \left(x\right)}^{b \left(x\right)} \frac{\partial}{\partial x} \left[f \left(x , t\right)\right] \mathrm{dt} + f \left(x , b \left(x\right)\right) \frac{\mathrm{db}}{\mathrm{dx}} - f \left(x , a \left(x\right)\right) \frac{\mathrm{da}}{\mathrm{dx}}$

In this case, we have:

$\frac{d}{\mathrm{dx}} \left[{\int}_{x}^{{x}^{2}} {t}^{3} \mathrm{dt}\right]$,

where $a \left(x\right) = x$, $b \left(x\right) = {x}^{2}$, and $f \left(x , t\right) = {t}^{3}$.

We obtain, by plugging in $a \left(x\right)$ and $b \left(x\right)$ in place of $t$ for the $f \left(x , a \left(x\right)\right)$ and $f \left(x , b \left(x\right)\right)$ terms:

$= {\cancel{{\int}_{x}^{{x}^{2}} 0 \mathrm{dt}}}^{0} + \stackrel{| \left[{t}^{3}\right] {|}_{t = {x}^{2}}}{\overbrace{{\left({x}^{2}\right)}^{3}}} \cdot \textcolor{red}{\frac{\mathrm{db}}{\mathrm{dx}}} - \stackrel{| \left[{t}^{3}\right] {|}_{t = x}}{\overbrace{{\left(x\right)}^{3}}} \cdot \textcolor{red}{\frac{\mathrm{da}}{\mathrm{dx}}}$

since the derivative of ${t}^{3}$ with respect to $x$ is $0$ ($t$ does not vary when $x$ varies), and the integral of $0$ is $0$ (a flat horizontal line along the $x$ axis has no area).

We then take the derivative of $b \left(x\right) = x$ with respect to $x$ and $a \left(x\right) = {x}^{2}$ with respect to $x$ to get:

$\implies \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[{\int}_{x}^{{x}^{2}} {t}^{3} \mathrm{dt}\right]} = {x}^{6} \cdot \textcolor{red}{2 x} - {x}^{3} \cdot \textcolor{red}{1}$

$= \textcolor{b l u e}{2 {x}^{7} - {x}^{3}}$