← Back to learning objective

Setting up the right-endpoint Riemann sum and limit

Consider the function f(x)=x3f(x) = x^3 on the interval [0,2][0, 2]. We want to express the area under the curve using a right-endpoint Riemann sum and then write the exact area as a limit.

Divide [0,2][0,2] 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 definite integral ∫02x3 dx\displaystyle\int_0^2 x^3\,dx as lim⁡n→∞∑i=1nf(xi∗) Δx\displaystyle\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\,\Delta x, substituting your expressions for f(xi∗)f(x_i^*) and Δx\Delta x.

State the full limit expression (you do not need to evaluate the sum or the limit).

Hint

Recall that for [a,b][a,b] divided 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)=x3f(x) = x^3.

Step-by-step solution

Since [0,2][0,2] is divided into nn equal subintervals, the width of each subinterval is Δx=2−0n=2n\Delta x = \frac{2-0}{n} = \frac{2}{n}. The right endpoint of the ii-th subinterval starting from a=0a=0 is xi∗=0+iΔx=2inx_i^* = 0 + i\Delta x = \frac{2i}{n}.

Substituting into f(x)=x3f(x) = x^3 gives f(xi∗)=(2in)3=8i3n3f(x_i^*) = \left(\frac{2i}{n}\right)^3 = \frac{8i^3}{n^3}. Combining with Δx=2n\Delta x = \frac{2}{n}, each term of the Riemann sum is 8i3n3⋅2n=16i3n4\frac{8i^3}{n^3}\cdot\frac{2}{n} = \frac{16i^3}{n^4}. Therefore ∫02x3 dx=lim⁡n→∞∑i=1n16i3n4\displaystyle\int_0^2 x^3\,dx = \lim_{n\to\infty}\sum_{i=1}^n \frac{16i^3}{n^4}.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap