Problem 73 - Calculus Early Transcendentals: Pearson New International Edition

Question

first recognize the given limit as a definite integral and then evaluate that integral by the Second Fundamental Theorem of Calculus. $$ \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{3 i}{n}\right)^{2} \frac{3}{n} $$

Step-by-Step Solution

Answer

The limit evaluates to 3.

Step 1: Recognize as a Riemann Sum

The given limit represents a Riemann sum. Riemann sums approximate the integral of a function over a given interval. Here, examine the expression within the sum: \( \left(\frac{3i}{n}\right)^2 \frac{3}{n} \). This suggests the function \( f(x) = (3x)^2=9x^2 \) and the interval approximated as \([0, 1]\) since \( \frac{3i}{n} \) represents subinterval points.

Step 2: Identify the Function

From the Riemann sum expression, identify the function being integrated. The inner part \((\frac{3i}{n})^2\) corresponds to \(f(x) = 9x^2\), and the differential element is \(\Delta x = \frac{3}{n}\). The function being integrated, therefore, is \(9x^2\) over the interval \([0, 1]\).

Step 3: Use the Definite Integral Formulation

Now translate the Riemann sum into its definite integral form: \( \lim_{n \to \infty} \sum_{i=1}^{n} 9 \left(\frac{3i}{n}\right)^2 \frac{3}{n} = \int_{0}^{1} 9x^2 \, dx \). This transformation converts the sum to a limit integral over the interval \([0, 1]\).

Step 4: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus allows us to evaluate the integral \( \int_{0}^{1} 9x^2 \, dx \). This is done by finding an antiderivative of \(9x^2\), which is \(3x^3\). Now, evaluate this antiderivative at the upper limit and subtract its evaluation at the lower limit: \( \left[3x^3\right]_{0}^{1} \).

Step 5: Evaluate the Integral

Substitute the limits of integration into the antiderivative: \(3(1)^3 - 3(0)^3 = 3 \). This calculation provides the value of the original limit in integral form.

Key Concepts

Riemann Sum

The Riemann Sum is a fundamental concept in calculus that helps us approximate the area under a curve. This approximation allows us to express limits of sums as definite integrals. Visualize it as slicing a shape into thin vertical strips, calculating the area of each strip, and then adding them up.

• In the expression \( \sum_{i=1}^{n}\left(\frac{3 i}{n}\right)^{2} \frac{3}{n}\), we have a Riemann sum approximation.


• The function \(f(x) = 9x^2\) represents the shape under which we sum these areas.


• The interval we are concerned with here is \([0, 1]\).

Riemann sums are powerful because, as you increase the number of slices (\(n \to \infty\)), the approximation becomes more accurate, eventually matching the exact area indicated by the definite integral. This is why recognizing such patterns as Riemann sums is crucial for moving into integral calculus.

Second Fundamental Theorem of Calculus

This theorem connects differentiation and integration, forming the backbone of integral calculus. It lets us find definite integrals using antiderivatives. In simple terms, it says that if you have a continuous function \(f\), you can find the accumulated area (the integral) easily through its antiderivative, \(F\).

• The theorem states: \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).


• Here, \(F\) is any function whose derivative is \(f\), and \([a, b]\) is your interval.

In practice, you don't need to sum an infinite number of slices to find an area. As in our exercise, you compute an antiderivative of \(9x^2\), which is \(3x^3\), and evaluate it at the interval's boundaries (0 to 1). This simplifies our calculation of the original Riemann sum to just a subtraction \( F(b) - F(a) \).

Antiderivative

An antiderivative, or indefinite integral, of a given function is a new function whose derivative is the original function. It's like undoing the derivative to find the original function that was differentiated.

• For our function \(9x^2\), its antiderivative is \(3x^3\).


• The process to find an antiderivative involves reversing the power rule of differentiation: if \(f(x) = ax^n\), then its antiderivative is \(\frac{a}{n+1}x^{n+1} + C\), where \(C\) is a constant.

Antiderivatives are crucial when applying the Second Fundamental Theorem of Calculus. They provide the means to evaluate the definite integral by allowing subtraction of boundary evaluations. In our exercise, evaluating \(3x^3\) at the bounds \([0, 1]\) gives us the area under the curve, completing our calculation of the integral. Understanding this concept helps simplify complex integration tasks, making them more approachable and manageable.