# How do you find the derivative of #G(x)=int (t^2-3t)dt# from #[1,x^2]#?

# G'(x) = 2x^5-6x^3 #

We have:

If we let

Then from the first part of the Fundamental Theorem of Calculus, we have:

And from the second part of the Fundamental Theorem of Calculus, we have:

Therefore,

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To find the derivative of ( G(x) = \int_{1}^{x^2} (t^2 - 3t) dt ), we can use the Fundamental Theorem of Calculus and the chain rule.

First, we find the antiderivative of ( t^2 - 3t ), which is ( \frac{1}{3}t^3 - \frac{3}{2}t^2 ).

Next, we evaluate this antiderivative at the upper and lower limits of integration and subtract.

So, ( G(x) = \left[ \frac{1}{3}t^3 - \frac{3}{2}t^2 \right]_{1}^{x^2} )

Evaluating at the upper limit, we get ( \frac{1}{3}x^6 - \frac{3}{2}x^4 ).

Evaluating at the lower limit, we get ( \frac{1}{3} - \frac{3}{2} ).

Subtracting these results, we get ( G(x) = \frac{1}{3}x^6 - \frac{3}{2}x^4 - \frac{1}{3} + \frac{3}{2} ).

Now, we differentiate ( G(x) ) with respect to ( x ) to find the derivative:

( G'(x) = \frac{d}{dx} \left( \frac{1}{3}x^6 - \frac{3}{2}x^4 - \frac{1}{3} + \frac{3}{2} \right) )

( G'(x) = 2x^5 - 6x^3 ).

So, the derivative of ( G(x) ) is ( 2x^5 - 6x^3 ).

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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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