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

Question

Charles Arocha · Accepted Answer

Find a formula for the Riemann Sum with #n# equal intervals and then evaluate tha limit as #n rarr oo#. The limit you are looking for is the integral, so find #int_0^1 (x^2 +1) dx# So, if you have it available to you, use the Fundamental Theorem of Calculus: #int_0^1 (x^2 +1) dx = 1/3x^3+x]_0^1# # = (1/3 +1 ) - 0+0) = 4/3#

Elijah Abram · Answer

undefined 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} $.