# How do I find the riemann sum of #y = x^2 + 1# for [0,1] at infinitely small intervals?

So, if you have it available to you, use the Fundamental Theorem of Calculus:

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To find the Riemann sum of ( y = x^2 + 1 ) for the interval ([0,1]) at infinitely small intervals, you would use the definite integral. The Riemann sum approaches the definite integral as the width of the intervals approaches zero.

The definite integral of ( y = x^2 + 1 ) over the interval ([0,1]) is calculated as follows:

[ \int_{0}^{1} (x^2 + 1) , dx ]

[ = \left[ \frac{x^3}{3} + x \right]_{0}^{1} ]

[ = \left( \frac{1^3}{3} + 1 \right) - \left( \frac{0^3}{3} + 0 \right) ]

[ = \left( \frac{1}{3} + 1 \right) - 0 ]

[ = \frac{4}{3} ]

Therefore, the Riemann sum of ( y = x^2 + 1 ) for the interval ([0,1]) at infinitely small intervals is ( \frac{4}{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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