Chapter 11: Problem 41
Use any basic integration formula or formulas to find the indefinite integral. State which integration formula(s) you used to find the integral. $$ \int e^{x} \sqrt{1-e^{x}} d x $$
Short Answer
Expert verified
The indefinite integral \(\int e^{x} \sqrt{1-e^{x}} dx = -2/3 (1-e^{x})^{3/2} + C\) where \(C\) is the integration constant. Integration formula used during this solution is the power rule of integration.
Step by step solution
01
Identify the substitution
To solve this integral, we will use the method of substitution. In this case, due to the presence of \(e^{x}\) and \(\sqrt{1-e^{x}}\), the selected substitution is \(u = e^{x}\). After differentiating both sides with respect to \(x\), we obtain \(du/dx = e^{x}\), therefore \(dx = du/e^{x}\) or \(dx = du/u\).
02
Substitute values
Substitute \(u\) in the integral, resulting in \(\int u \sqrt{1-u} du/u\), which simplifies to \(\int \sqrt{1-u} du\). This form is much simpler to handle.
03
Solve the Simplified Integral
The integral \(\int \sqrt{1-u} du\) is a basic form of integral that can be solved easily using power rule of integration which states: \(\int x^n dx = x^{n+1}/(n+1) + c\). Here \(n = -1/2\). Solving this we get \(-2/3 (1-u)^{3/2}\) as the indefinite integral of \(\sqrt{1-u}\).
04
Substitute back the original variable
Substitute \(u = e^{x}\) back into the expression to reveal the original variable, which leads to \(-2/3 (1-e^{x})^{3/2}\).
05
Add Integration Constant
Add integration constant at the end as the integral obtained is an indefinite integral. Therefore, the final answer is \(-2/3 (1-e^{x})^{3/2} + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
One effective method for solving complex integrals is through integration by substitution, which simplifies expressions by changing variables. This technique is particularly useful when dealing with compositions of functions, like a function nested inside another function.
Let's look at how substitution works: first, you identify a portion of the integral as a new variable, usually denoted by 'u'. This portion should make the integral simpler once rewritten in terms of 'u'. The next step involves differentiating this new variable 'u' with respect to 'x' to find 'du/dx', which provides the substitute for 'dx'. Finally, the integral is expressed in 'u', making it easier to solve. After integrating with respect to 'u', you substitute back to the original variable, 'x'.
For the given exercise, the substitution was chosen as (u = e^{x}), transforming the integral into a simpler form and leveraging the power rule for integration, which we'll discuss next.
Let's look at how substitution works: first, you identify a portion of the integral as a new variable, usually denoted by 'u'. This portion should make the integral simpler once rewritten in terms of 'u'. The next step involves differentiating this new variable 'u' with respect to 'x' to find 'du/dx', which provides the substitute for 'dx'. Finally, the integral is expressed in 'u', making it easier to solve. After integrating with respect to 'u', you substitute back to the original variable, 'x'.
For the given exercise, the substitution was chosen as (u = e^{x}), transforming the integral into a simpler form and leveraging the power rule for integration, which we'll discuss next.
Power Rule of Integration
The power rule of integration is a fundamental technique used frequently to solve integrals. It states that the integral of (x^n dx) is (x^{n+1}/(n+1) + C), as long as n ≠ -1. The constant 'C' is the integration constant, denoting that there are an infinite number of antiderivatives.
In our exercise, the power rule applies after simplification through substitution. The term (sqrt{1-u}) is rewritten as (1-u)^{1/2}, and we treat this as (u^n) with n = -1/2 to apply the power rule. This results in -2/3 (1-u)^{3/2}, seamlessly integrating the expression. Knowing this rule can immensely simplify many integral problems, so it's a crucial formula to remember and understand in calculus.
In our exercise, the power rule applies after simplification through substitution. The term (sqrt{1-u}) is rewritten as (1-u)^{1/2}, and we treat this as (u^n) with n = -1/2 to apply the power rule. This results in -2/3 (1-u)^{3/2}, seamlessly integrating the expression. Knowing this rule can immensely simplify many integral problems, so it's a crucial formula to remember and understand in calculus.
Integration Formulas
Calculus is full of integration formulas that can be utilized to compute integrals. From straight-forward power functions to exponential, logarithmic, and trigonometric functions, each type of function has its own set of integration rules.
Following the correct formula is essential for reaching the correct solution. In our example, once the variable was substituted, we were working with a square root function. Recognizing that you can rewrite square roots as powers allowed us to use the power integration formula.
To enhance your problem-solving skills, familiarize yourself with a variety of these formulas, as they act as tools in a mathematician's toolbox. You never know which one you'll need to solve a new problem!
Following the correct formula is essential for reaching the correct solution. In our example, once the variable was substituted, we were working with a square root function. Recognizing that you can rewrite square roots as powers allowed us to use the power integration formula.
To enhance your problem-solving skills, familiarize yourself with a variety of these formulas, as they act as tools in a mathematician's toolbox. You never know which one you'll need to solve a new problem!
Integration Constant
When you find the indefinite integral of a function, you're actually finding a family of functions, all of which are correct. This family is represented by adding an integration constant, usually denoted as 'C', to the result.
This constant accounts for the fact that differentiating a constant gives zero, so any constant can be present in the antiderivative without affecting differentiation. The presence of 'C' encapsulates all possible vertical shifts of the antiderivative.
As seen in step 5 of our solution, after substituting back to the original variable, we added 'C' to the result to convey that there is not just a single solution, but an infinite set of antiderivatives. In summary, never forget to add 'C' when computing indefinite integrals, as it's an essential part of the correct answer.
This constant accounts for the fact that differentiating a constant gives zero, so any constant can be present in the antiderivative without affecting differentiation. The presence of 'C' encapsulates all possible vertical shifts of the antiderivative.
As seen in step 5 of our solution, after substituting back to the original variable, we added 'C' to the result to convey that there is not just a single solution, but an infinite set of antiderivatives. In summary, never forget to add 'C' when computing indefinite integrals, as it's an essential part of the correct answer.