Question

Multiple Choice If the average value of the function \(f\) on the interval \([ a , b ]\) is \(10 ,\) then \(\int _ { a } ^ { b } f ( x ) d x =\) (A) \(\frac { 10 } { b - a } \quad\) (B) \(\frac { f ( a ) + f ( b ) } { 10 } \quad\) (C) \(10 b - 10 a\) \(( \mathbf { D } ) \frac { b - a } { 10 } \quad ( \mathbf { E } ) \frac { f ( b ) + f ( a ) } { 20 }\)

Short answer

The correct answer to the problem is (C), since \(\int _ { a } ^ { b } f ( x ) d x = 10(b - a)\).

Step-by-step solution

Understand the formula for average value of a function

The average value \(f_{avg}\) of a function \(f(x)\) over the interval [a, b] is given by the formula \(f_{avg} = \frac{\int _ { a } ^ { b } f ( x ) d x}{b - a}\).

Substitute the given average value

Given that the average value of the function \(f\) on the interval [a, b] is 10, substitute 10 in for \(f_{avg}\) in the formula: 10 = \frac{\int _ { a } ^ { b } f ( x ) d x}{b - a}

Solve for the integral

To solve for the integral \(\int _ { a } ^ { b } f ( x ) d x\), we multiply both sides of the equation from Step 2 by \(b - a\). This gives us \(\int _ { a } ^ { b } f ( x ) d x = 10(b - a) \).

Key concepts

Definite Integrals

Understanding definite integrals is crucial for students who are working with the average value of functions. A definite integral represents the accumulation of quantities, which can be areas under a curve, or in this particular context, it enables us to determine the total sum of function values across an interval.

For a function, say f(x), the definite integral from a to b is denoted as \[ \int _ { a } ^ { b } f ( x ) d x \] and represents the net area bounded by the function, the x-axis, and the vertical lines x=a and x=b. When calculating the average value of a function over an interval, the definite integral plays a pivotal role as it essentially sums up the continuous values that f(x) takes on from a to b. Once evaluated, it can be divided by the length of the interval to find the average value.

Definite integrals also obey certain properties, such as the additive property, which can be utilized when dealing with multiple integrals over adjacent intervals. Knowledge of these properties helps in simplifying complex integral expressions.

Function Integration

The process of evaluating a definite integral is referred to as function integration. It involves finding a function, F(x), known as the antiderivative, whose derivative is the function f(x) under consideration. When we find this antiderivative, we can use it to evaluate the definite integral by applying the Fundamental Theorem of Calculus.

According to this theorem, if F(x) is an antiderivative of f(x), then: \[ \int _ { a } ^ { b } f ( x ) d x = F(b) - F(a) \] Function integration is not always straightforward, as it requires a familiarity with integration techniques which may include u-substitution, integration by parts, or recognizing patterns that match standard integrals. Mastering these techniques is vital for effectively solving calculus problems, including those involving the average value of functions.

Calculus Problem Solving

Calculus problem solving involves a systematic approach to tackle questions by applying calculus concepts and methods, such as differentiation and integration. When solving an average value problem, it's important to systematically follow steps like those outlined in the original exercise.

The steps begin with understanding the concept and formula involved—in this case, the formula for the average value of a function. From there, we substitute known values into the formula and solve for the unknown, which is typically the definite integral in these types of problems.

Effective problem solving in calculus also necessitates the recognition of patterns, simplifying expressions before integrating, and verifying solutions, which may include checking results using geometric interpretations or using derivative rules as confirmation.