Skip to main content
Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.4.53a

Average value with a parameter Consider the function ฦ’(๐“) = a๐“ (1โ€•๐“) on the interval [0, 1], where a is a positive real number.
(a) Find the average value of ฦ’ as a function of a .

Verified step by step guidance
1
Step 1: Recall the formula for the average value of a function ฦ’(๐“) on the interval [a, b], which is given by: 1(b-a)fxdx. In this case, the interval is [0, 1] and the function is ฦ’(๐“) = a๐“(1 - ๐“).
Step 2: Substitute the interval [0, 1] and the function ฦ’(๐“) = a๐“(1 - ๐“) into the formula for average value: 110^1(ax(1-x))dx. This simplifies to: 0^1ax(1-x)dx.
Step 3: Expand the integrand a๐“(1 - ๐“) to simplify the integral. This becomes: a(x-x2). The integral now looks like: a0^1(x-x2)dx.
Step 4: Break the integral into two separate parts: a(0^1xdx-0^1x2dx). Compute each integral separately: 0^1xdx and 0^1x2dx. Use the power rule for integration: xndx=xn+1n+1.
Step 5: After computing the integrals, combine the results and multiply by the constant 'a' to find the average value of ฦ’(๐“) as a function of 'a'. The final expression will represent the average value of the function over the interval [0, 1].

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Value of a Function

The average value of a continuous function ฦ’ over an interval [a, b] is calculated using the formula (1/(b-a)) * โˆซ[a to b] ฦ’(x) dx. This concept is essential for determining how the function behaves on the specified interval, providing a single representative value that summarizes the function's overall trend.
Recommended video:
06:37
Average Value of a Function

Definite Integral

A definite integral represents the accumulation of quantities, such as area under a curve, over a specific interval. In this context, it is used to compute the integral of the function ฦ’(x) = a๐“(1 - ๐“) from 0 to 1, which is necessary for finding the average value of the function.
Recommended video:
05:43
Definition of the Definite Integral

Parameter in Functions

A parameter is a variable that influences the behavior of a function but is not the primary variable of interest. In this case, 'a' is a parameter that affects the shape and scale of the function ฦ’(x) = a๐“(1 - ๐“), and understanding its role is crucial for expressing the average value as a function of 'a'.
Recommended video:
Guided course
05:59
Eliminating the Parameter
Related Practice
Textbook Question

Bounds on an integral Suppose ฦ’ is continuous on [a, b] with ฦ’''(๐“) > 0 on the interval. It can be shown that (bโ€•a) ฦ’ [(a + b) /2] โ‰ค โˆซโ‚แต‡ ฦ’(๐“) d๐“ โ‰ค (bโ€•a) [ (ฦ’(a) + ฦ’(b)) /2]                                                         

                                                                                                                                                                               

(a) Assuming ฦ’ is nonnegative on [a, b], draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b. 

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ฦ’, ฦ’', and ฦ’'' are continuous functions for all real numbers.                                                                                                                                                           

                                                                                                                                                                    

(a) โˆซ ฦ’(๐“) ฦ’'(๐“) d๐“ = ยฝ (ฦ’(๐“))ยฒ + C.

1
views
Textbook Question

Mass from density A thin 10-cm rod is made of an alloy whose density varies along its length according to the function shown in the figure. Assume density is measured in units of g/cm. In Chapter 6, we show that the mass of the rod is the area under the density curve.

(a) Find the mass of the left half of the rod (0 โ‰ค x โ‰ค 5) .

1
views
Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(a) Consider the linear function ฦ’(๐“) = 2x + 5 and the region bounded by its graph and the x-axis on the interval [3,6]. Suppose the area of this region is approximated using midpoint Riemann sums. Then the approximations give the exact area of the region for any number of subintervals.

1
views
Textbook Question

Area functions for constant functions Consider the following functions ฦ’ and real numbers a (see figure).

(a) Find and graph the area function A(๐“) = โˆซโ‚หฃ ฦ’(t) dt for ฦ’.

ฦ’(t) = 5 , a = 0

1
views
Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(a) If ฦ’ is a constant function on the interval [a,b], then the right and left Riemann sums give the exact value of โˆซโ‚แต‡ ฦ’(๐“) d๐“, for any positive integer n.