How do you use the Fundamental Theorem of Calculus to find the derivative of #int sqrt(3t+ sqrt(t)) dt# from 5 to tanx?

Question

Ezekiel Alexander · Accepted Answer

#= sec^2 x \ sqrt(3tan x + sqrt(tan x)) #

FTC, Pt 1: if #F(u)= int _a^u \ f(t)\ dt #, then #F'(u)=f(u) #.

With u(x) = tan x, and #F = F(u(x))#, it follows from the chain rule that:

#(dF)/(dx) = F'(u) \ u'(x)#

So:

#d/(dx) ( int_5^(tan(x)) sqrt(3t+ sqrt(t)) \ \ dt) #

#= sqrt(3tan x + sqrt(tan x)) \ d/(dx) (tan x ) #

#= sec^2 x \ sqrt(3tan x + sqrt(tan x)) #

Ezra Adams · Answer

undefined To find the derivative of the integral $ \int_{5}^{\tan(x)} \sqrt{3t + \sqrt{t}} \, dt $, we can use the Fundamental Theorem of Calculus, which states that if $ f(x) $ is continuous on $[a, b]$ and $ F(x) $ is an antiderivative of $ f(x) $ on $[a, b]$, then

$$ \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) $$

Given the integral $ \int_{5}^{\tan(x)} \sqrt{3t + \sqrt{t}} \, dt $, we first need to find an antiderivative of the integrand. Let's denote $ u = 3t + \sqrt{t} $, then $ du = (3 + \frac{1}{2\sqrt{t}}) dt $. Solving for $ dt $, we get $ dt = \frac{du}{(3 + \frac{1}{2\sqrt{t}})} $. Substituting this back into the integral, we get:

$$ \int_{5}^{\tan(x)} \sqrt{3t + \sqrt{t}} \, dt = \int_{u(5)}^{u(\tan(x))} \sqrt{u} \cdot \frac{du}{(3 + \frac{1}{2\sqrt{t}})} $$

Now, we can find the antiderivative of $ \sqrt{u} \cdot \frac{du}{(3 + \frac{1}{2\sqrt{t}})} $ with respect to $ u $. After finding the antiderivative, we can differentiate it with respect to $ x $ to find the derivative.