/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 Suppose that \( f(1) = 2 \), \( ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 71

Suppose that \( f(1) = 2 \), \( f(4) = 7 \), \( f^\prime(1) = 5 \), \( f^\prime(4) = 3 \) and \( f^{\prime\prime} \) is continuous. Find the value of \( \displaystyle \int_1^4 x f^{\prime\prime} (x)\ dx \).

Short Answer

Expert verified
The value of \( \int_1^4 x f''(x) \, dx \) is 2.

Step by step solution

01

Understand the Problem

The problem asks us to find the integral \( \int_1^4 x f''(x)\, dx \) given specific values of \( f(x) \) and its derivatives at certain points. We know \( f(1) = 2 \), \( f(4) = 7 \), \( f'(1) = 5 \), and \( f'(4) = 3 \). Moreover, \( f'' \) is continuous.
02

Integrate by Parts

The method of integration by parts states that for any functions \( u(x) \) and \( v(x) \), \( \int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) u'(x) \, dx \). For this problem, let \( u(x) = x \) and \( dv = f''(x) \, dx \).
03

Determine Parts for Integration by Parts

If \( u(x) = x \), then \( du = dx \). If \( dv = f''(x) \, dx \), then \( v = f'(x) \). Substituting these into the integration by parts formula gives us: \[ \int_1^4 x f''(x) \, dx = \left[ x f'(x) \right]_1^4 - \int_1^4 f'(x) \, dx \].
04

Evaluate \( x f'(x) \) at the bounds

We need to evaluate \( x f'(x) \) at the bounds of integration. At \( x=4 \), we have \( 4 \cdot f'(4) = 4 \times 3 = 12 \). At \( x=1 \), \( 1 \cdot f'(1) = 1 \times 5 = 5 \). So, \( \left[ x f'(x) \right]_1^4 = 12 - 5 = 7 \).
05

Evaluate \( \int_1^4 f'(x) \, dx \)

The integral \( \int_1^4 f'(x) \, dx \) can be evaluated as \( f(x) \) from 1 to 4, which means \( \left[ f(x) \right]_1^4 = f(4) - f(1) = 7 - 2 = 5 \).
06

Combine Results

From Step 3, we have \( \int_1^4 x f''(x) \, dx = \left[ x f'(x) \right]_1^4 - \int_1^4 f'(x) \, dx \). Substituting in the results from Steps 4 and 5, we get: \( 7 - 5 = 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics focused on change. It deals with rates of change and accumulations. There are two main parts: differential calculus and integral calculus.
  • Differential calculus focuses on the concept of derivatives. This helps in understanding how a quantity changes with respect to another.
  • Integral calculus is about accumulation. It involves adding up small quantities to determine a total value, typically using integrals.
In the context of this exercise, we deal with a particular form of integration technique known as integration by parts. This technique breaks an integral of a product of functions into simpler parts. In rough terms, it finds the integral from two functions multiplied together.
Definite Integrals
Definite integrals calculate the area under a curve for a specific interval. It helps find the total accumulation within that range. In this problem, we're evaluating a definite integral from 1 to 4.
  • The notation for a definite integral is \( \int_a^b f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.
  • The result of a definite integral is a numerical value, which represents the net area between the curve and the x-axis over a given interval.
We applied integration by parts to evaluate the integral of \( \int_1^4 x f''(x) \, dx \). By solving the integral, the aim was to express this complexity into recognizable parts, leading to an easier computation of the net area between the bounds.
Second Derivative
The second derivative of a function, represented as \( f''(x) \), gives insight into the concavity of a function. It is the derivative of the first derivative \( f'(x) \).
  • While the first derivative indicates the rate of change, the second derivative shows the rate at which this rate of change is shifting.
  • A positive second derivative means the function is concave up, and a negative value indicates it is concave down.
In our exercise, the integral of the second derivative with multiplication by the variable \( x \) is evaluated, demonstrating how the behavior of \( f''(x) \) interacts with the variable within a given interval. This application of the second derivative through integration by parts helps simplify complex expressions to enhance comprehensibility and solve practical problems.

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Most popular questions from this chapter

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_0^5 \frac{1}{\sqrt[3]{5 - x}}\ dx \)

Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson's Rule to approximate the given integral with the specified value of \( n \). (Round your answers to six decimal places.) \( \displaystyle \int_0^4 x^3 \sin x\ dx \) , \( n = 8 \)

Use the Comparison Theorem to determine whether the integral is convergent or divergent. \( \displaystyle \int_1^\infty \frac{x + 1}{\sqrt{x^4 - x}}\ dx \)

Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster from the observed (two-dimensional) density that can be analyzed from a photograph. Suppose that in a spherical cluster of radius \( R \) the density of stars depends only on the distance \( r \) from the center of the cluster. If the perceived star density is given by \( y(s) \), where s is the observed planar distance from the center of the cluster, and \( x(r) \) is the actual density, it can be shown that $$ y(s) = \int_s^R \frac{2r}{\sqrt{r^2 - s^2}} x(r)\ dr $$ If the actual density of stars in a cluster is \( x(r) = \frac{1}{2} (R - r)^2 \), find the perceived density \( y(s) \).

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_1^\infty \frac{\ln x}{x}\ dx \)
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