Problem 71
Question
Derive the formulas \(\int e^{x} d x=e^{x}+c\) and \(\int e^{-x} d x=-e^{-x}+c\)
Step-by-Step Solution
Verified Answer
\(\int e^{x} d x=e^{x}+c\) and \(\int e^{-x} d x=-e^{-x}+c\)
1Step 1: Derive Integral of \(e^x\)
Start by finding the integral of \(e^{x}\). The function e^x is its own derivative and its own integral. Antidifferentiating \(e^x\) with respect to x gives: \(\int e^{x} d x=e^{x}+c\) .
2Step 2: Derive Integral of \(e^{-x}\)
Now find the integral of \(e^{-x}\). This is a bit trickier, because the exponent is negative. You can solve this by using the substitution method: let \(u = -x\), then \(du = -dx\). The integral now becomes: \(-\int e^{u} du\), which is \(-e^{u}+c\). Replace \(u\) by \(-x\) to get the final answer: \(-e^{-x}+c\).
Key Concepts
AntiderivativesExponential FunctionsSubstitution Method
Antiderivatives
An antiderivative, also known as an indefinite integral, is a fundamental concept in calculus. It represents the reverse process of differentiation. When you find the antiderivative of a function, you are essentially determining the original function before it was differentiated.
For example, when dealing with the exponential function \(e^x\), its antiderivative is \(e^x + c\). The \(c\) represents an arbitrary constant, showing that there are infinitely many antiderivatives for a function, each differing by a constant value. This concept highlights how differentiation strips away the constant, while integration brings it back.
Antiderivatives are crucial because they help in solving various practical problems, including areas under curves and in differential equations analysis. They are notations of accumulation, adding up infinitely small quantities to derive a larger, often more meaningful total.
For example, when dealing with the exponential function \(e^x\), its antiderivative is \(e^x + c\). The \(c\) represents an arbitrary constant, showing that there are infinitely many antiderivatives for a function, each differing by a constant value. This concept highlights how differentiation strips away the constant, while integration brings it back.
Antiderivatives are crucial because they help in solving various practical problems, including areas under curves and in differential equations analysis. They are notations of accumulation, adding up infinitely small quantities to derive a larger, often more meaningful total.
Exponential Functions
Exponential functions have unique properties that make them intriguing and essential in calculus. The standard form of these functions is \(f(x) = a \, e^{bx}\), where \(a\) and \(b\) are constants, and \(e\) is the base of the natural logarithm, approximately equal to 2.718.
A particularly interesting aspect of exponential functions, especially \(e^x\), is that they are their own derivatives and integrals. This means that differentiating \(e^x\) with respect to \(x\) gives \(e^x\), and integrating \(e^x\) gives \(e^x + c\). This property simplifies many computations in calculus and provides an elegant symmetry to the functions.
A particularly interesting aspect of exponential functions, especially \(e^x\), is that they are their own derivatives and integrals. This means that differentiating \(e^x\) with respect to \(x\) gives \(e^x\), and integrating \(e^x\) gives \(e^x + c\). This property simplifies many computations in calculus and provides an elegant symmetry to the functions.
- They grow rapidly: Exponential functions exhibit exponential growth, meaning they increase in value at a consistently accelerating rate.
- Applications: They are used to model phenomena in physics, finance, biology, and many other fields.
Substitution Method
The substitution method is a useful technique in calculus to simplify the process of integration, especially when dealing with complex functions. It works by changing the variable of integration to make the integral easier to evaluate.
Consider the function \(e^{-x}\). Direct integration can be complicated, but by substituting \(u = -x\), the problem becomes more manageable. According to the substitution method, you also need \(du = -dx\). Hence, the integral \(\int e^{-x} dx\) becomes \(-\int e^{u} du\), which simplifies directly to \(-e^{u} + c\). Finally, by substituting back the original variable, you return to \(-e^{-x} + c\).
Consider the function \(e^{-x}\). Direct integration can be complicated, but by substituting \(u = -x\), the problem becomes more manageable. According to the substitution method, you also need \(du = -dx\). Hence, the integral \(\int e^{-x} dx\) becomes \(-\int e^{u} du\), which simplifies directly to \(-e^{u} + c\). Finally, by substituting back the original variable, you return to \(-e^{-x} + c\).
- Ease of Use: Substitution helps break down an integral into simpler parts.
- Widely Applicable: It is versatile for many types of problems in integration.
Other exercises in this chapter
Problem 70
Derive the formulas \(\int \sec ^{2} x d x=\tan x+c\) and \(\int \sec x \tan x \, d x=\sec x+c\)
View solution Problem 71
Identify the integrals to which the Fundamental Theorem of Calculus applies; the other integrals are called improper integrals. (a) \(\int_{0}^{4} \frac{1}{x-4}
View solution Problem 72
Identify the integrals to which the Fundamental Theorem of Calculus applies; the other integrals are called improper integrals. (a) \(\int_{0}^{1} \frac{1}{\sqr
View solution Problem 73
Identify each sum as a Riemann sum and evaluate the limit. (a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\left[\sin \frac{\pi}{n}+\sin \frac{2 \pi}{n}+\cdots+\s
View solution