Question

Find the value(s) of \(c\) guaranteed by the Mean Value Theorem for Integrals for the function over the indicated interval. $$ f(x)=10-2^{x}, \quad[0,3] $$

Short answer

The value of \(c\) guaranteed by the Mean Value Theorem for Integrals for the given function over the interval \([0, 3]\) is \(c \approx 1.63093\).

Step-by-step solution

Compute the integral

First, evaluate the definite integral from 0 to 3 for the given function. This is done by applying integral properties and formulas as follows:\[ \int_0^3 (10-2^{x}) \, dx = \left[10x - \frac{2^{x}}{\ln(2)}\right]_0^3 \] which results in \(30 - \frac{8}{\ln(2)}\).

Apply the Mean Value Theorem for Integrals

According to the theorem, there must be a value \(c\) in the interval \(0 \leq c \leq 3\) such that \[(b-a)f(c) = \int_a^b f(x) \, dx\]Substitute the given upper and lower limits into the left hand side of the equation, as follows:\[(3-0)(10-2^c) = 30 - \frac{8}{\ln2}\]which simplifies to \[30 - 3 \cdot 2^c = 30 - \frac{8}{\ln2}\]

Solve for \(c\)

By simplifying the previous equation, we get\[ 3 \cdot 2^c = \frac{8}{\ln2}\]Solving this for \(c\) gives\[ c = \frac{\ln\left(\frac{8}{3\ln2}\right)}{\ln2}\]which simplifies to \[c \approx 1.63093\]

Key concepts

Definite Integral

The concept of the definite integral is a fundamental piece of the calculus puzzle, providing a way to calculate the accumulation of quantities or the area under a curve. When you see an expression like \[ \int_a^b f(x) \, dx \], you're looking at the definite integral of the function \(f(x)\) over the interval from \(a\) to \(b\).

The process involves evaluating the function's antiderivative at the upper and lower bounds of the interval and then finding the difference. For instance, in our exercise, the antiderivative of \(10-2^x\) was calculated and then evaluated at the bounds 0 and 3. The result is the net area between the function and the x-axis from \(x=0\) to \(x=3\), providing the total 'accumulated value' over this interval.

This is a crucial concept in all fields that involve quantities changing over time, such as physics, economics, and biology. In essence, through definite integration, we're able to make precise the idea of 'total change' across an interval.

Exponential Functions

Turning our attention to exponential functions, these are functions of the form \(f(x) = a^x\), where \(a\) is a positive constant. They are exceptional because the rate of change, or derivative, of an exponential function is proportional to the value of the function itself — a property that is unique in mathematics.

In our exercise, we deal with the function \(2^x\), which is a classic example of an exponential function. Exponential functions like this one grow rapidly and are typically used to model processes that increase or decrease at rates proportional to their size, such as population growth, radioactive decay, and interest in finance.

The Integral of Exponential Functions

Computing the integral of an exponential function involves its own anti-derivative, which introduces the concept of the natural logarithm, denoted as \(\ln (x)\). This connects us seamlessly to another critical area in calculus — logarithmic functions.

Logarithmic Properties

Logarithmic functions are the inverse of exponential functions. If you see an expression like \(y = \ln(x)\), this translates to the statement that \(e^y = x\), where \(e\) is the natural logarithm base, an irrational constant approximately equal to 2.71828. One of the most important logarithmic properties used in calculus is that the logarithm of a power, such as \(\ln(a^b)\), is equal to the exponent times the logarithm of the base: \(b\cdot\ln(a)\).

In the case of solving for \(c\) in our exercise, we've encountered the logarithmic property when isolating \(c\) in the equation. Simplifying the equation with logarithms allows us to find the value of \(c\) that makes the statement true in the context of the Mean Value Theorem for Integrals.

This relationship between exponential and logarithmic functions is vital for solving equations in calculus where variables are in exponents — as in our example — and allows us to handle the complexity of growth and decay processes in the natural and social sciences.