Question

Explain why \(\frac{1}{100} \sum_{i=1}^{100} f\left(\frac{i}{100}\right)\) is an estimate of \(\int_{0}^{1} f(x) d x\)

Short answer

Explain why the summation \(\frac{1}{100}\sum_{i=1}^{100} f\left(\frac{i}{100}\right)\) is an estimate of the definite integral \(\int_{0}^{1} f(x) d x\). The summation given can be identified as a right-endpoint Riemann sum with 100 equal sub-intervals for the function \(f(x)\) on the interval \([0, 1]\). When the function is continuous and smooth, a Riemann sum serves as a reasonable estimate of the definite integral. Although in this case, the Riemann sum is calculated using only 100 sub-intervals, it still provides a helpful approximation for the definite integral, which is why the summation is an estimate of the definite integral \(\int_{0}^{1} f(x) d x\).

Step-by-step solution

Understanding Definite Integral as Limit of Riemann Sums

A definite integral, written as \(\int_{a}^{b} f(x) d x\), represents the signed area between the curve of a function \(f(x)\) and the \(x\)-axis on the interval \([a, b]\). We can approximate this area by dividing the interval into sub-intervals and summing the areas of rectangles inscribed within the curve. This is precisely a Riemann sum. The Riemann sum is given as: \(\sum_{i=1}^{n} f(x_i^*)\Delta x_i\), where \(x_i^*\) is a point in the i-th sub-interval \([x_{i-1}, x_i]\) and \(\Delta x_i = x_i - x_{i-1}\). The definite integral can be expressed as the limit of a Riemann sum as the number of sub-intervals (\(n\)) approaches infinity: \(\int_{a}^{b} f(x) d x = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i^*)\Delta x_i\).

Relating the Given Summation to Riemann Sum

To connect the given summation to the Riemann sum, consider partitioning the interval \([0, 1]\) into \(100\) equal sub-intervals. This gives us a partition width of \(\Delta x = \frac{1}{100}\). Now, let us find the right-endpoint Riemann sum for this partition. The right-endpoint Riemann sum has \(x_i^* = \frac{i}{100}\) for each sub-interval: \(\sum_{i=1}^{100} f\left(\frac{i}{100}\right)\Delta x = \sum_{i=1}^{100} f\left(\frac{i}{100}\right)\frac{1}{100}\) This expression is the same as the given summation: \(\frac{1}{100}\sum_{i=1}^{100} f\left(\frac{i}{100}\right)\).

Explaining the Connection between the Summation and the Integral

Having established that the given summation precisely matches the right-endpoint Riemann sum with \(100\) equal subintervals (using right endpoints) for the function \(f(x)\) on the interval \([0, 1]\), we can now connect it to the definite integral. As the number of sub-intervals increases, the Riemann sum approximation of the definite integral becomes more accurate. Although the given summation represents an approximation using only 100 subintervals, it still serves as a reasonable estimate of the definite integral on the interval \([0,1]\): \(\int_{0}^{1} f(x) d x \approx \frac{1}{100}\sum_{i=1}^{100} f\left(\frac{i}{100}\right)\) This relationship is especially true when the function \(f(x)\) is continuous and smooth on the interval \([0, 1]\), which would allow the Riemann sum to converge more quickly to the definite integral value. So, the given summation is an estimate of the definite integral \(\int_{0}^{1} f(x) d x\).