Question

In Exercises \(43-48,\) use the limit process to find the area of the region between the graph of the function and the \(x\) -axis over the given interval. Sketch the region. $$ y=x^{2}+1, \quad[0,3] $$

Short answer

The area under the curve for the function \(y=x^{2}+1\) over the interval [0,3] is 9 square units.

Step-by-step solution

Setting up the Riemann Sum

Start by dividing the region from 0 to 3 into n equal intervals, each of width Δx. Each interval is represented by \(x_{i} = i * Δx\) where \(i = 0, 1, . . ., n-1, n\) and \(Δx = \frac{b-a}{n} = \frac{3-0}{n} = \frac{3}{n}\). Hence, the Riemann Sum is given by: \(S_n = \sum_{i=1}^{n} f(x_i) * Δx\).

Writing the limit of the Riemann Sum

Substitute \(x_i\) and \(Δx\) into \(S_n\), this gives: \(S_n = \sum_{i=1}^{n} f(i * Δx) * Δx = \sum_{i=1}^{n} [(i * Δx)^2 + 1] * Δx\). The integral will be the limit of this sum as n approaches infinity: \(\int_{0}^{3} (x^2 + 1) dx = \lim_{n->∞} S_n\).

Evaluating the limit

To solve the limit, expand the summation in the current expression for \(S_n\) and find the limit as \(n\) approaches infinity, this gives: \(\lim_{n->∞} \sum_{i=1}^{n} [(i * Δx)^2 + 1] * Δx = \lim_{n->∞} \sum_{i=1}^{n} [(i * \frac{3}{n})^2 + 1] * \frac{3}{n}\). The limit of this sum can be calculated directly once the formula for the sum of the first \(n\) squares is recalled: \(\sum_{i=1}^{n} i^2 = \frac{n*(n+1)(2n+1)}{6}\). So, \( \lim_{n->∞} [(n*(n+1)(2n+1) *18)/6*n^2 + 3n]/n = \lim_{n->∞} [3(2n+1)/2 + 3] = 9\).

Finishing the problem

Calculate the area of the curve to get: \(Area = \int_{0}^{3} (x^2 + 1) dx = 9 square units\). The process is complete.