How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #g(x) = int(1 / (t^3 + 1)) dt# from [1,x]?

Answer 1

The theorem says:

Given the function:

#y=int_(h(x))^g(x)f(t)dt#

then:

#y'=f(g(x))*g'(x)-f(h(x))*h'(x)#.

So:

#g'(x)=1/(x^3+1)*1-1/(1^3+1)*0=1/(x^3+1)#.
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Answer 2

To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function ( g(x) = \int_{1}^{x} \frac{1}{t^3 + 1} , dt ), we differentiate the integral expression with respect to ( x ). According to the theorem, if ( F(x) = \int_{a}^{x} f(t) , dt ), then ( F'(x) = f(x) ).

So, in this case, the derivative of ( g(x) ) is ( g'(x) = \frac{1}{x^3 + 1} ).

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Answer from HIX Tutor

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.

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.

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