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Chapter 7: Problem 48

Find a positive value of \(k\) such that the area under the graph of \(y=e^{2 x}\) over the interval \([0, k]\) is 3 square units.

Short Answer

Expert verified
The positive value of \(k\) is \(k = \frac{\ln(7)}{2}\).

Step by step solution

01

Write the Integral Expression

To find the area under the curve, we need to integrate the function \(y = e^{2x}\) over the interval \([0, k]\). The integral expression for the area \(A\) is: \[ A = \int_0^k e^{2x} \, dx \]
02

Integrate the Function

Find the antiderivative of \(e^{2x}\). Using substitution, let \(u = 2x\), then \(du = 2 \, dx\), which implies that \(dx = \frac{1}{2} du\). The integral becomes:\[ \int e^{u} \frac{1}{2} \, du = \frac{1}{2} e^u \]Substitute back \(u = 2x\) to get:\[ \frac{1}{2} e^{2x} \]
03

Apply Limits of Integration

Apply the limits of integration to the antiderivative from 0 to \(k\):\[ \left[ \frac{1}{2} e^{2x} \right]_0^k = \frac{1}{2} e^{2k} - \frac{1}{2} e^{0} \]Simplify it:\[ \frac{1}{2} e^{2k} - \frac{1}{2} \]
04

Set the Integral Equal to Given Area

Set the expression for the area equal to 3 square units:\[ \frac{1}{2} e^{2k} - \frac{1}{2} = 3 \]
05

Solve for k

Solve \(\frac{1}{2} e^{2k} - \frac{1}{2} = 3 \). First add \(\frac{1}{2}\) to both sides:\[ \frac{1}{2} e^{2k} = 3.5 \]Then multiply both sides by 2:\[ e^{2k} = 7 \]Take the natural logarithm on both sides to solve for \(k\):\[ 2k = \ln(7) \]Finally, divide by 2:\[ k = \frac{\ln(7)}{2} \]
06

Interpret the Solution

The positive value of \(k\) such that the area under the curve from \(x = 0\) to \(x = k\) is 3 square units is:\[ k = \frac{\ln(7)}{2} \]

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Key Concepts

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

Exponential Function
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. A basic form of an exponential function is given as \( y = a^{x} \), where \( a \) is a positive real number, and \( x \) is the exponent. However, one of the most significant exponential functions in calculus is \( y = e^{x} \), where \( e \) is Euler's number, approximately 2.71828. Key characteristics of exponential functions include:
  • They have a constant rate of growth or decay, depending on the sign of the exponent.
  • The function \( y = e^{x} \) is always positive and extends upwards as \( x \) increases.
  • The function is continuous and differentiable everywhere on the real line.
Understanding exponential functions is crucial when calculating areas under curves, as they exhibit exponential growth, which is important in various fields such as finance, natural sciences, and engineering.
Integration by Substitution
Integration by substitution, sometimes considered the 'reverse' of the chain rule, is a technique used to simplify finding antiderivatives. It is especially useful when dealing with composite functions, such as exponential functions with linear terms in the exponent.Here's how it works:
  • Identify a part of the integrand that can be substituted to simplify the integral. In our case, for \( y = e^{2x} \), let \( u = 2x \).
  • Determine the derivative of \( u \), \( du = 2 \, dx \), and solve for \( dx \): \( dx = \frac{1}{2} du \).
  • Rewrite the integral in terms of \( u \), giving \( \int e^{u} \frac{1}{2} \, du \).
  • Integrate the simpler expression and then substitute back for \( x \).
This method transforms a complex integral into a much simpler one, allowing easier computation of definite integrals.
Limits of Integration
When calculating definite integrals, the limits of integration specify the interval over which you integrate, defining the segments of the area under the curve. In this exercise, the limits are from \( 0 \) to \( k \). Key aspects are:
  • After finding the antiderivative, you apply these limits to determine the net area.
  • For \( \int_0^k e^{2x} \, dx \), the result is \( \frac{1}{2} e^{2k} - \frac{1}{2} e^{0} \).
  • Evaluating the antiderivative at the upper limit \( k \) and subtracting the value of the function evaluated at the lower limit \( 0 \) yields the desired area.
The limits of integration help solve practical problems by defining precise intervals, which is key when determining exact quantities like areas or total values.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is the inverse function of the exponential function \( e^{x} \). It answers the question "to what power must we raise \( e \) to obtain \( x \)?" Some essential properties include:
  • \( \ln(e) = 1 \), because \( e^1 = e \).
  • \( \ln(1) = 0 \), since any number raised to the power of 0 is 1.
  • It is only defined for positive values of \( x \).
In the exercise, taking the natural logarithm of both sides helps solve exponential equations. When solving \( e^{2k} = 7 \), taking \( \ln \) on both sides simplifies finding \( k \), because it allows the exponent to be brought down, resulting in \( 2k = \ln(7) \). This manipulation transforms complex exponential equations into simpler algebraic ones.

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Most popular questions from this chapter

Let the velocity function for a particle that is at the origin initially and moves along an \(s\) -axis be \(v(t)=0.5-t e^{-t}\). (a) Generate the velocity versus time curve, and use it to make a conjecture about the sign of the displacement over the time interval \(0 \leq t \leq 5\) (b) Use a CAS to find the displacement.

(a) Use a graphing utility to generate the graph of $$f(x)=\frac{1}{100}(x+2)(x+1)(x-3)(x-5) $$ and use the graph to make a conjecture about the sign of the integral $$\int_{-2}^{5} f(x) d x$$ (b) Check your conjecture by evaluating the integral.

Let the velocity function for a particle that is at the origin nitially and moves along an \(s\) -axis be \(v(t)=t \ln (t+0.1)\). (a) Generate the velocity versus time curve, and use it to make a conjecture about the sign of the displacement over the time interval \(0 \leq t \leq 1\) (b) Use a CAS to find the displacement.

Use Part 2 of the Fundamental Theorem of Calculus to find the derivative. $$\frac{d}{d x} \int_{x}^{0} \frac{t}{\cos t} d t$$ [Hint: Use Definition \(7.5 .3(\mathrm{b}) .]\)

It was shown in the proof of the Mean-Value Theorem for Integrals that if \(f\) is continuous on \([a, b],\) and if \(m \leq f(x) \leq M\) on \([a, b],\) then $$m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a)$$Isee \((8)] .\) These inequalities make it possible to obtain bounds on the size of a definite integral from bounds on the size of its integrand. This is illustrated. Find the maximum and minimum values of \(\sqrt{x^{3}+2}\) for \(0 \leq x \leq 3,\) and use these values to find bounds on the value of the integral $$\int_{0}^{3} \sqrt{x^{3}+2} d x$$
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