Problem 67
Question
Suppose that \(f\) is a positive function whose graph over the interval \([a, b]\) lies under the chord joining \((a, f(a))\) to \((b,\) \(f(b))\). By calculating the area of the trapezoid with vertices \((a, 0),(b, 0),(b, f(b)),\) and \((a, f(a)),\) deduce that $$ \int_{a}^{b} f(x) d x \leq(b-a) \cdot\left(\frac{f(a)+f(b)}{2}\right) $$.
Step-by-Step Solution
Verified Answer
The area of the trapezoid formed by the points (a, 0), (b, 0), (b, f(b)), and (a, f(a)) is greater than or equal to the area under the curve of f(x) over [a, b], thus \(\int_a^b f(x) \, dx \leq (b-a) \cdot \left(\frac{f(a) + f(b)}{2}\right)\).
1Step 1: Area of the trapezoid
The trapezoid formed by \((a, f(a))\) and \((b, f(b))\) with the x-axis has area \(\frac{(b-a)[f(a)+f(b)]}{2}\).
2Step 2: Compare with integral
Since the graph lies under the chord (function is concave down), the trapezoid overestimates the area: \(\int_a^b f(x)\,dx \leq \frac{(b-a)[f(a)+f(b)]}{2}\).
Key Concepts
Trapezoid RuleInequalityIntegral Approximation
Trapezoid Rule
The Trapezoid Rule is a numerical technique used to approximate the value of a definite integral. This method converts the process of integration, which can be complex, into a simpler form of calculating the area of a trapezoid. Imagine having a graph of a function over an interval
- The x-axis serves as the base of the trapezoid, extending from point \( a \) to \( b \).
- The heights of the trapezoid at these points are the values of the function \( f(a) \) and \( f(b) \).
- The top of the trapezoid is created by connecting these two function values with a straight line.
Inequality
In mathematics, an inequality shows the relationship between two values that are not equal, indicating which one is larger or smaller. Inequality is a central theme when dealing with integrals and areas, frequently used to compare them as seen in this exercise.For the given problem, we were asked to show that the area under the curve \( \int_{a}^{b} f(x) \, dx \) is less than or equal to the area of the trapezoid formed by the line segment connecting \((a, f(a))\) to \((b, f(b))\). This is expressed as:\[\int_{a}^{b} f(x) \, dx \leq (b-a) \cdot \left(\frac{f(a)+f(b)}{2}\right)\]The statement holds because the function's graph lies below the line segment joining the points. Essentially, the inequality indicates that the curve's area is less than or at most equal to that of the trapezoid. This comparison is a straightforward application of how the integration can provide a lower bound for areas bounded by more complex shapes.
Integral Approximation
Integral approximation is an essential part of numerical mathematics, especially when dealing with functions that are difficult to integrate analytically. There are various methods for approximating definite integrals, each offering a balance between simplicity and accuracy.
Descriptive understanding of approximation:
- Methods like the Trapezoid Rule provide simple approximations that are easy to compute.
- Approximations are especially useful for functions that do not have straightforward antiderivatives.
- They take advantage of the properties of basic shapes like rectangles or trapezoids to estimate the area under the curve.
Other exercises in this chapter
Problem 67
A sum of integrals of the form \(\int_{a}^{b} f(x) d x\) is given. Express the sum as a single integral of form \(\int_{c}^{d} g(y) d y\). \(\int_{0}^{\sqrt{2}
View solution Problem 67
Calculate the integrals. $$ \int \frac{\arctan (x)}{1+x^{2}} d x $$
View solution Problem 68
A definite integral is given. Do not attempt to calculate its value \(V .\) Instead, find the extreme values of the integrand on the interval of integration, an
View solution Problem 68
Calculate the integrals. $$ \int \frac{\exp (x)}{1+\exp (x)} d x $$
View solution