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Setting up the right-endpoint Riemann sum and its limit

Consider the function f(x)=x2f(x) = x^2 on the interval [0,3][0, 3].

Divide [0,3][0,3] into nn subintervals of equal width, and let xi∗x_i^* be the right endpoint of the ii-th subinterval.

  1. Find Δx\Delta x in terms of nn.
  2. Find xi∗x_i^* in terms of ii and nn.
  3. Write the Riemann sum ∑i=1nf(xi∗)Δx\sum_{i=1}^{n} f(x_i^*)\Delta x explicitly in terms of ii and nn (you do not need to simplify or evaluate the sum).
  4. Write the definite integral ∫03x2 dx\displaystyle\int_0^3 x^2\,dx as the limit of this Riemann sum.

Give your final answer as the limit expression lim⁡n→∞∑i=1nf(xi∗)Δx\displaystyle\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x with f(xi∗)Δxf(x_i^*)\Delta x written out explicitly in terms of ii and nn.

Hint

Recall that for an interval [a,b][a,b] split into nn equal pieces, Δx=b−an\Delta x = \frac{b-a}{n} and the right endpoint of the ii-th subinterval is xi∗=a+iΔxx_i^* = a + i\Delta x. Substitute these into f(x)=x2f(x) = x^2.

Step-by-step solution

Since [a,b]=[0,3][a,b] = [0,3], the width of each subinterval is Δx=3−0n=3n\Delta x = \frac{3-0}{n} = \frac{3}{n}. The right endpoint of the ii-th subinterval is xi∗=0+iΔx=3inx_i^* = 0 + i\Delta x = \frac{3i}{n}.

Substituting into f(x)=x2f(x) = x^2 gives f(xi∗)=(3in)2=9i2n2f(x_i^*) = \left(\frac{3i}{n}\right)^2 = \frac{9i^2}{n^2}, so f(xi∗)Δx=9i2n2⋅3n=27i2n3f(x_i^*)\Delta x = \frac{9i^2}{n^2}\cdot\frac{3}{n} = \frac{27i^2}{n^3}.

Therefore the definite integral is expressed as ∫03x2 dx=lim⁡n→∞∑i=1n27i2n3.\int_0^3 x^2\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{27i^2}{n^3}.

Behavior of delta x as n grows

Hint
Step-by-step solution

Delta x equals three divided by n, so as n grows larger the value of delta x gets smaller. Thus delta x decreases toward zero as n approaches infinity. This shrinking width is exactly what allows the Riemann sum to converge to the exact area under the curve.

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