Problem 31 - Applied Calculus

Question

(a) Use a calculator or computer to find \(\int_{0}^{6}\left(x^{2}+1\right) d x\) Represent this value as the area under a curve. (b) Estimate \(\int_{0}^{6}\left(x^{2}+1\right) d x\) using a left-hand sum with \(n=3 .\) Represent this sum graphically on a sketch of \(f(x)=x^{2}+1 .\) Is this sum an overestimate or underestimate of the true value found in part (a)? (c) Estimate \(\int_{0}^{6}\left(x^{2}+1\right) d x\) using a right-hand sum with \(n=3 .\) Represent this sum on your sketch. Is this sum an overestimate or underestimate?

Step-by-Step Solution

Answer

(a) Integral is 78. (b) Left-hand sum is 46, an underestimate. (c) Right-hand sum is 118, an overestimate.

Step 1: Calculate Definite Integral

Use a calculator or software to compute the definite integral \( \int_{0}^{6} (x^2 + 1) \, dx \). The integral of \( x^2 + 1 \) is found using the antiderivative \( F(x) = \frac{x^3}{3} + x \). Evaluate this from 0 to 6. So, \[ \int_{0}^{6} (x^2 + 1) \, dx = \left[ \frac{6^3}{3} + 6 \right] - \left[ \frac{0^3}{3} + 0 \right] = 72 + 6 = 78. \]This integral value is the area under the curve \( x^2 + 1 \) from \( x=0 \) to \( x=6 \).

Step 2: Use Left-hand Sum (n=3)

To estimate the integral with a left-hand sum, divide the interval \([0, 6]\) into 3 equal subintervals: \([0, 2], [2, 4], [4, 6]\). The width of each subinterval is 2.Evaluate function values at the left endpoints: \( f(0), f(2), f(4) \).\[ f(0) = 0^2 + 1 = 1, \quad f(2) = 2^2 + 1 = 5, \quad f(4) = 4^2 + 1 = 17. \]The left-hand sum is:\[ 2( f(0) + f(2) + f(4) ) = 2(1 + 5 + 17) = 46. \]Since the function \( x^2 + 1 \) is increasing, the left-hand sum gives an underestimate of the actual area.

Step 3: Use Right-hand Sum (n=3)

For the right-hand sum, use the right endpoints of the subintervals \([0, 2], [2, 4], [4, 6]\), which means taking \( f(2), f(4), f(6) \).\[ f(2) = 5, \quad f(4) = 17, \quad f(6) = 37. \]The right-hand sum is:\[ 2( f(2) + f(4) + f(6) ) = 2(5 + 17 + 37) = 118. \]The function \( x^2 + 1 \) is increasing, so the right-hand sum is an overestimate of the actual integral value.

Key Concepts

Area Under a Curve

When discussing integrals and their applications, "area under a curve" is a fundamental concept that helps us find the total accumulation. Imagine a curve on a graph, above the x-axis, such as the function defined by \( f(x) = x^2 + 1 \). The \( ext{definite integral} \) \( \int_{0}^{6} (x^2 + 1) \, dx \) helps establish the size of the area beneath this curve from \( x = 0 \) to \( x = 6 \). This area represents the total accumulation of values for the given function over the interval.To be precise, by calculating the definite integral, we determine this area mathematically, and for \( f(x) = x^2 + 1 \), it evaluates to 78. This value indicates how tall and wide the section of the graph is, below the curve, within the specified domain. Like shading a real plot, this mathematical result showcases growth, and is crucial in various real-world applications such as in determining distance, probability, and economic variables.

Left-Hand Sum

The "left-hand sum" is a numerical estimation method to approximate the area under a curve, using rectangles. When using a left-hand sum, the height of each rectangle is determined by the value of the function at the left endpoint of subintervals. Here's how it works: divide the interval \( [0, 6] \) into equal parts. With \( n = 3 \), we have subintervals \( [0, 2], [2, 4], [4, 6] \). For each subinterval, calculate its height at the left endpoint:

• At \( x=0 \), \( f(0)=1 \)
• At \( x=2 \), \( f(2)=5 \)
• At \( x=4 \), \( f(4)=17 \)

Multiply these values by the width (2) and sum them up to get an approximate total area: \( 2(1 + 5 + 17) = 46 \). Since the function \( f(x) = x^2 + 1 \) increases, rectangles sit below the curve, suggesting a left-hand sum results in an underestimate of the actual area beneath the curve.

Right-Hand Sum

Similarly to the left-hand sum, the "right-hand sum" estimates the area under a curve using rectangles. However, here the rectangles' height is determined by the function's value at the right endpoint of each subinterval. In practice, partition the interval \( [0, 6] \) into \( n = 3 \) parts, yielding intervals \( [0, 2], [2, 4], [4, 6] \).

• At \( x=2 \), \( f(2)=5 \)
• At \( x=4 \), \( f(4)=17 \)
• At \( x=6 \), \( f(6)=37 \)

Calculate the right endpoints, multiply each by the width (2), and sum them to approximate the area: \( 2(5 + 17 + 37) = 118 \).Since \( f(x) = x^2 + 1 \) is an increasing function, the right-hand sum is often an overestimate, as the rectangles include area beyond the curve itself.

Overestimate and Underestimate

In calculus, approximations using sums often relate to whether they overestimate or underestimate the true area under a curve. Knowing when each occurs allows you to manage expectations regarding the accuracy of your estimate.

• Underestimate: This happens in left-hand sums for increasing functions, as the top of each rectangle lies strictly below the curve.
• Overestimate: Conversely, a right-hand sum will often overestimate because the rectangles extend above the curve, capturing extra area not truly under it.

Recognizing these tendencies is especially vital when you estimate integrals manually, as it clarifies potential errors. The goal is not just to approximate but to consciously adjust, if greater precision is needed, by deploying finer subdivisions or using adjustment techniques like the trapezoidal rule. These considerations guide practical decisions in real-world applications where such integrals play pivotal roles.