(a) For integer $n \geq 1$, use upper and lower Riemann sums with $n$ equal subdivisions to find an upper and lower bound for the value of:

$\int_0^1 (1 - 2x^3) \, dx$

You may use the following fact without proof: $\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2 (n+1)^2}{4}$

(b) Evaluate: $\int_0^1 (1 - 2x^3) \, dx$ using the definition of the Riemann integral.

Neetesh Kumar | October 21, 2024

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Riemann Sum
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Step-by-step solution:

Part (a): Upper and Lower Riemann Sums

We are asked to approximate the integral $\int_0^1 (1 - 2x^3) \, dx$ using Riemann sums. First, divide the interval $[0,1]$ into $n$ equal subintervals, each of width $\Delta x = \frac{1}{n}$.

The function $f(x) = 1 - 2x^3$ is decreasing on the interval $[0,1]$, which means that for the upper sum, we will evaluate the function at the left endpoint of each subinterval, and for the lower sum, we will evaluate it at the right endpoint.

Step 1: Lower Sum

The lower sum is calculated by evaluating the function at the right endpoints of the subintervals:

$x_k = \frac{k}{n}$, where $k = 1, 2, 3, \ldots, n$

The lower sum is:

$L_n = \sum_{k=1}^{n} f \left( \frac{k}{n} \right) \Delta x = \sum_{k=1}^{n} \left( 1 - 2 \left( \frac{k}{n} \right)^3 \right) \frac{1}{n}$

Simplifying the expression:

$L_n = \frac{1}{n} \sum_{k=1}^{n} \left( 1 - 2 \frac{k^3}{n^3} \right)$

$L_n = \frac{1}{n} \left( \sum_{k=1}^{n} 1 - 2 \frac{1}{n^3} \sum_{k=1}^{n} k^3 \right)$

Using the given fact for $\sum_{k=1}^{n} k^3$:

$\sum_{k=1}^{n} k^3 = \frac{n^2 (n+1)^2}{4}$

Substituting this into the expression for $L_n$:

$L_n = \frac{1}{n} \left( n - 2 \frac{1}{n^3} \cdot \frac{n^2 (n+1)^2}{4} \right)$

Simplifying:

$L_n = 1 - \frac{(n+1)^2}{2n^2}$

Step 2: Upper Sum

The upper sum is calculated by evaluating the function at the left endpoints of the subintervals:

$x_{k-1} = \frac{k-1}{n}$, where $k = 1, 2, 3, \ldots, n$

The upper sum is:

$U_n = \sum_{k=1}^{n} f \left( \frac{k-1}{n} \right) \Delta x = \sum_{k=1}^{n} \left( 1 - 2 \left( \frac{k-1}{n} \right)^3 \right) \frac{1}{n}$

Simplifying the expression:

$U_n = \frac{1}{n} \sum_{k=1}^{n} \left( 1 - 2 \frac{(k-1)^3}{n^3} \right)$

$U_n = \frac{1}{n} \left( \sum_{k=1}^{n} 1 - 2 \frac{1}{n^3} \sum_{k=1}^{n} (k-1)^3 \right)$

Since $\sum_{k=1}^{n} (k-1)^3 = \sum_{k=0}^{n-1} k^3$, we use the same formula for the sum of cubes:

$\sum_{k=0}^{n-1} k^3 = \frac{(n-1)^2 n^2}{4}$

Substituting into $U_n$:

$U_n = 1 - \frac{n^2}{2(n-1)^2}$

Step 3: Conclusion for Part (a)

The lower sum provides a lower bound and the upper sum provides an upper bound for the integral:

$\text{Lower bound: } \bigg(1 - \frac{(n+1)^2}{2n^2}\bigg)$

$\text{Upper bound: } \bigg(1 - \frac{n^2}{2(n-1)^2}\bigg)$

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Part (b): Evaluate the Integral

We now evaluate the integral $\int_0^1 (1 - 2x^3) \, dx$ using the definition of the Riemann integral.

The integral is:

$\int_0^1 (1 - 2x^3) \, dx = \int_0^1 1 \, dx - 2 \int_0^1 x^3 \, dx$

First, evaluate $\int_0^1 1 \, dx$:

$\int_0^1 1 \, dx = x \Big|_0^1 = 1$

Next, evaluate $\int_0^1 x^3 \, dx$:

$\int_0^1 x^3 \, dx = \frac{x^4}{4} \Big|_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$

Now, substitute these into the original expression:

$\int_0^1 (1 - 2x^3) \, dx = 1 - 2 \times \frac{1}{4} = 1 - \frac{1}{2} = \frac{1}{2}$

Final Answer:

(a) The upper and lower bounds are:

• Lower bound: $\bigg(1 - \dfrac{(n+1)^2}{2n^2}\bigg)$
• Upper bound: $\bigg(1 - \dfrac{n^2}{2(n-1)^2}\bigg)$

(b) The exact value of the integral is: $\int_0^1 (1 - 2x^3) \, dx = \dfrac{1}{2}$

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