How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #h(x)= sqrt(3+r^3) dr# for a=1, #b=x^2#?

Question

Christopher Allers · Accepted Answer

I wonder if perhaps you mean to ask how to use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #h(x)= int_1^(x^2)  sqrt(3+r^3) dr#  If so, we'll need the Fundamental Theorem Part 1 and also the chain rule. #d/(du)(int_1^(u)  sqrt(3+r^3) dr) = sqrt(3+u^3)#, but we've been asked for #d/(dx)(h(x)# so we'll use the chain rule with #u = x^2# #d/(dx)(int_1^(u)  sqrt(3+r^3) dr) = sqrt(3+u^3) * ((du)/(dx)) = 2xsqrt(3+x^6)#,

Axel Alvarez · Answer

undefined To find the derivative of the function $ h(x) = \int_{1}^{x^2} \sqrt{3 + r^3} \, dr $, you can use Part 1 of the Fundamental Theorem of Calculus, which states that if $ f $ is continuous on $[a, b]$ and $ F $ is an antiderivative of $ f $ on $[a, b]$, then

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

For the given function $ h(x) $, we need to evaluate its derivative with respect to $ x $. So, apply the Fundamental Theorem of Calculus to $ h(x) $ as follows:

$$ h'(x) = \frac{d}{dx} \left( \int_{1}^{x^2} \sqrt{3 + r^3} \, dr \right) $$

Now, to find $ h'(x) $, differentiate the integral with respect to $ x $ using the Chain Rule:

$$ h'(x) = \frac{d}{dx} \left( \int_{1}^{x^2} \sqrt{3 + r^3} \, dr \right) = \frac{d}{dx} \left( F(x^2) - F(1) \right) $$

$$ = \frac{d}{dx} F(x^2) = f(x^2) \cdot 2x $$

Where $ f(r) = \sqrt{3 + r^3} $ and $ F(r) $ is an antiderivative of $ f(r) $.

So, $ h'(x) = 2x \sqrt{3 + (x^2)^3} $.