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Magazine Chapter 5: Problem 62
Evaluate. $$ \int_{0}^{b} 2 e^{-2 x} d x $$
Short Answer
Expert verified
The value of the integral is \(1 - e^{-2b}\).
Step by step solution
01
Understand the Integral
The given integral is \(\text{鈭珆_{0}^{b} 2 e^{-2x} \, dx\). This represents the area under the curve of the function \(\text{2e}^{-2x}\) from \(x=0\) to \(x=b\).
02
Find the Antiderivative
To solve the integral, we need to find the antiderivative of \(2 e^{-2x}\). \( \text{鈭珆 2 e^{-2x} \, dx \)
03
Use U-Substitution
Let \(u = -2x\). Then \(du = -2 \, dx\rightarrow dx = -\frac{1}{2} \, du\). Substituting these into the integral, we get \(\text{鈭珆 2 e^u \left(-\frac{1}{2}\right) du = -\text{鈭珆 e^u du = -e^u + C\)
04
Substitute Back
After integrating, we need to revert back to the variable \x\. The antiderivative becomes \(-e^{-2x} + C\).
05
Evaluate the Definite Integral
Evaluate \(-e^{-2x}\) from \(0\) to \(b\)\. This means calculating the difference \([\text{-e}^{-2b} - \text{-e}^{0} ]\).
06
Simplify the Expression
Since \(e^0 = 1\), the expression becomes \([-e^{-2b} + 1] = 1 - e^{-2b}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
An antiderivative is a function that 'reverses' the process of differentiation.
When we differentiate a function, we find its derivative. The antiderivative does the opposite 鈥� it finds a function whose derivative is the given function.
For instance, if the derivative of a function is given by \(f'(x) = 2 e^{-2x} \), we want to find a function \(F(x)\) such that \(F'(x) = 2 e^{-2x}\). In this case, the antiderivative of \(2 e^{-2x}\) is \(-e^{-2x}\).
When we differentiate a function, we find its derivative. The antiderivative does the opposite 鈥� it finds a function whose derivative is the given function.
For instance, if the derivative of a function is given by \(f'(x) = 2 e^{-2x} \), we want to find a function \(F(x)\) such that \(F'(x) = 2 e^{-2x}\). In this case, the antiderivative of \(2 e^{-2x}\) is \(-e^{-2x}\).
U-Substitution
U-substitution is a method used to simplify the process of finding the antiderivative.
It involves substituting a part of the integral with a new variable, typically denoted as \u\. Consider the integral \(\text{鈭珆 2 e^{-2x} \, dx\)
By letting \(u = -2x\), we transform the integral, making it easier to solve.
This simplifies to \(\text{鈭珆 2 e^u \, \frac{-1}{2} \, du = - \text{鈭珆 e^u \, du\)
We then integrate with respect to the new variable and substitute back the original variable once integrated.
It involves substituting a part of the integral with a new variable, typically denoted as \u\. Consider the integral \(\text{鈭珆 2 e^{-2x} \, dx\)
By letting \(u = -2x\), we transform the integral, making it easier to solve.
This simplifies to \(\text{鈭珆 2 e^u \, \frac{-1}{2} \, du = - \text{鈭珆 e^u \, du\)
We then integrate with respect to the new variable and substitute back the original variable once integrated.
Exponential Function
An exponential function is a mathematical function of the form \(f(x) = a^x\), where the base \(a\) is a positive real number, and \(x\) is the exponent.
One common exponential function is \(e^x\), where \(e\) is Euler's number, approximately equal to 2.71828.
In the integral given \(\text{鈭珆 2 e^{-2x} \, dx\), we are dealing with an exponential function where the exponent is \(-2x\).
Exponential functions play a crucial role in calculus, particularly in integration, because their derivatives and antiderivatives are straightforward to compute.
One common exponential function is \(e^x\), where \(e\) is Euler's number, approximately equal to 2.71828.
In the integral given \(\text{鈭珆 2 e^{-2x} \, dx\), we are dealing with an exponential function where the exponent is \(-2x\).
Exponential functions play a crucial role in calculus, particularly in integration, because their derivatives and antiderivatives are straightforward to compute.
Definite Integral
A definite integral represents the area under a curve within a given interval.
The notation \(\text{鈭珆_{a}^{b} f(x) \, dx\) indicates that we are finding the area under the curve \(f(x)\) from \x = a\ to \x = b\.
For the given exercise, the definite integral \(\text{鈭珆_{0}^{b} 2 e^{-2x} \, dx\) means we are looking for the area from \x = 0\ to some point \ x = b\. The solution involves computing the antiderivative and then evaluating it at the bounds.
The notation \(\text{鈭珆_{a}^{b} f(x) \, dx\) indicates that we are finding the area under the curve \(f(x)\) from \x = a\ to \x = b\.
For the given exercise, the definite integral \(\text{鈭珆_{0}^{b} 2 e^{-2x} \, dx\) means we are looking for the area from \x = 0\ to some point \ x = b\. The solution involves computing the antiderivative and then evaluating it at the bounds.
Integration Techniques
Integration techniques are methods used to solve integrals.
They include straightforward integration, u-substitution, integration by parts, and partial fractions, among others.
In our exercise, we used u-substitution to simplify the integral of \(2 e^{-2x}\).
Understanding various integration techniques like u-substitution is essential for tackling more complex integrals in calculus.
They include straightforward integration, u-substitution, integration by parts, and partial fractions, among others.
In our exercise, we used u-substitution to simplify the integral of \(2 e^{-2x}\).
Understanding various integration techniques like u-substitution is essential for tackling more complex integrals in calculus.
