# Find the derivative of int_0^(x^3e^x) \(t^3+3)^17 \ dt?

Jul 28, 2017

Because we need the chain rule.

#### Explanation:

$y = g \left(u\right) = {\int}_{0}^{u} {\left({t}^{3} + 3\right)}^{17} \mathrm{dt}$ has

$\frac{\mathrm{dy}}{\mathrm{du}} = {\left({u}^{3} + 3\right)}^{17}$

and

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

so

$g ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = {\left({u}^{3} + 3\right)}^{17} \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} {e}^{3}\right)$

Jul 28, 2017

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{x}^{3} {e}^{x}} \setminus {\left({t}^{3} + 3\right)}^{17} \setminus \mathrm{dt} = \left({x}^{3} {e}^{x} + 3 {x}^{2} {e}^{x}\right) {\left({\left({x}^{3} {e}^{x}\right)}^{3} + 3\right)}^{17}$

#### 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).

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{x}^{3} {e}^{x}} \setminus {\left({t}^{3} + 3\right)}^{17} \setminus \mathrm{dt}$ ..... [A}

(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 the product rule, and the chain rule. Let:

$u = {x}^{3} {e}^{x} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \left({x}^{3}\right) \left(\frac{d}{\mathrm{de}} ^ x\right) + \left(\frac{d}{\mathrm{dx}} {x}^{3}\right) \left({e}^{x}\right)$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = {x}^{3} {e}^{x} + 3 {x}^{2} {e}^{x}$

The substituting into the integral [A], and applying the chain rule, we get:

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{x}^{3} {e}^{x}} \setminus {\left({t}^{3} + 3\right)}^{17} \setminus \mathrm{dt} = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{du}} \setminus {\int}_{0}^{u} \setminus {\left({t}^{3} + 3\right)}^{17} \setminus \mathrm{dt}$

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

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{x}^{3} {e}^{x}} \setminus {\left({t}^{3} + 3\right)}^{17} \setminus \mathrm{dt} = \frac{\mathrm{du}}{\mathrm{dx}} {\left({u}^{3} + 3\right)}^{17}$

And restoring the initial substitution we get:

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{x}^{3} {e}^{x}} \setminus {\left({t}^{3} + 3\right)}^{17} \setminus \mathrm{dt} = \left({x}^{3} {e}^{x} + 3 {x}^{2} {e}^{x}\right) {\left({\left({x}^{3} {e}^{x}\right)}^{3} + 3\right)}^{17}$

Which is option (2) in the initial question.