Darnell and stephanie have competing refreshment

Ответ:

jessie8022

06.08.2021

If one of the boundaries in an integral is a function of x, the fundamental theorem of calculus still applies. you just end up using the chain rule if you differentiate the integral.for example, consider the following integral: $\displaystyle \int_{-2}^{x^3} (t \cos t) \, dt$ if we let the anti-derivative of the integrand (t cos t) be represented by a function f, then f'(t) = t cos t. by the fundamental theorem of calculus, the integral can be rewritten as follows: $\displaystyle \int_{-2}^{x^3} (t \cos t) \, dt = f(x^3) - f(-2)$ if we were to differentiate this integral expression with respect to x, we would have to use the chain rule $\begin{aligned} \frac{d}{dx} \int_{-2}^{x^3} (t \cos t) \, dt & = \frac{d}{dx} \big[ f(x^3) - f(-2) \big] \\ & = \underbrace{f'(x^3) \cdot \tfrac{d}{dx}\left[x^3\right]}_{\text{chain rule}} - {} 0 \end{aligned}$ since f'(t) = t cos t, then f'(x³) = x³ cos x³ (just normal chain rule procedure) $\begin{aligned} \frac{d}{dx} \int_{-2}^{x^3} (t \cos t) \, dt & = x^3 \cos \left( x^3 \right) \cdot 3x^2 \\ & = 3x^5 \cos x^3 \end{aligned}$ ======here's another example $\begin{aligned} \int_{\tan x}^{2} \frac{u^2}{1+u} \, du & = f(2) - f(\tan x) & & \left( \text{\footnotesize where $f'(u) = \tfrac{u^2}{1+u} $} \right) \\ \\ \frac{d}{dx} \int_{\tan x}^{2} \frac{u^2}{1+u} \, du & = \frac{d}{dx} \big[ f(2) - f(\tan x) \big] \\ & = 0 - f'(\tan x) \cdot \tfrac{d}{dx} [\tan x] \\ & = - \frac{(\tan x)^2}{1+\tan x} \cdot \sec^2 x \\ & = - \frac{\tan^2 x \sec^2 x}{1 + \tan x} \end{aligned}$

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question