# How do you evaluate the definite integral #int (3x^2-2x+1)dx# from [1,5]?

The answer is

Here, we use the integral

Therefore

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To evaluate the definite integral of ( \int_{1}^{5} (3x^2 - 2x + 1) , dx ), first, find the antiderivative of the function ( 3x^2 - 2x + 1 ), which is ( x^3 - x^2 + x ). Then, substitute the upper limit (5) and lower limit (1) into the antiderivative and subtract the result of substituting the lower limit from the result of substituting the upper limit.

So, ( \int_{1}^{5} (3x^2 - 2x + 1) , dx = \left[ x^3 - x^2 + x \right]_{1}^{5} = (5^3 - 5^2 + 5) - (1^3 - 1^2 + 1) ).

Calculating this gives you the value of the definite integral.

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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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