# 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#?

If so, we'll need the Fundamental Theorem Part 1 and also the chain rule.

By signing up, you agree to our Terms of Service and Privacy Policy

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} ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you integrate #(x^2+2x-1) / (x^3 - x)#?
- How do you use part 1 of the fundamental theorem of calculus to find the derivative of #g(x)= int cos(sqrt( t))dt# 6 to x?
- How do you find the integral of # x^3/(1+x^2)#?
- What is #int (-2x^3-x^2+x+2 ) / (2x^2- x +3 )#?
- How do you evaluate the definite integral #int (2+x)dx# from [0,4]?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7