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Chapter 11: Problem 71

Evaluate \(\int_{3}^{3}(1+\sqrt{x}) e^{-x} d x\)

Short Answer

Expert verified
The definite integral \(\int_{3}^{3}(1+\sqrt{x}) e^{-x} d x\) evaluates to 0 since the bounds are the same.

Step by step solution

01

Antiderivative

The first step is to find the antiderivative of the given integrand. The given integrand is \( (1+\sqrt{x}) e^{-x}\) To find the antiderivative, we can use integration by parts. Let's set: \(u = 1+\sqrt{x}\) \(d v = e^{-x} d x\) Now, differentiate u and integrate dv: \(d u = \frac{1}{2\sqrt{x}} d x\) \(v = -e^{-x}\) Now, using the integration by parts formula, we have: \(\int (1+\sqrt{x}) e^{-x} d x = uv - \int vu dx \)
02

Calculate the integral

Now, we need to evaluate the integral: \(\int (1+\sqrt{x}) e^{-x} d x = u v - \int vu d x \) \(\int (1+\sqrt{x}) e^{-x} d x= -(1+\sqrt{x}) e^{-x} - \int -\frac{e^{-x}}{2 \sqrt{x}} d x \) Now, we integrate the leftover term (This can be done using substitution method. Substituting \(t = \sqrt{x}\), we get): \(\int -\frac{e^{-x}}{2 \sqrt{x}} d x = 2 \int e^{-t^2} d t \) However, there isn't an elementary antiderivative for \(e^{-t^2}\), so we'll leave this as: \(F(x) = -(1+\sqrt{x}) e^{-x} - 2\int e^{-t^2} d t\)
03

Apply the Fundamental Theorem of Calculus

Now, we apply the Fundamental Theorem of Calculus using the antiderivative F(x), and evaluate the definite integral between the bounds \(x=3\) to \(x=3\): \(\int_{3}^{3}(1+\sqrt{x}) e^{-x} d x = F(3)-F(3)\) Since the bounds are the same, the value of the definite integral is simply 0. So, \(\int_{3}^{3}(1+\sqrt{x}) e^{-x} d x = 0\).

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Key Concepts

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

Integration by Parts
The Integration by Parts formula is a powerful method used to integrate products of functions. To use it, we identify two parts of the integrand: one to differentiate and the other to integrate. Here is the basic formula:
  • \[ \int u \, dv = uv - \int v \, du \]
In our problem, we set:
  • \( u = 1 + \sqrt{x} \), meaning we'll differentiate this part.
  • \( dv = e^{-x} \, dx \), so we integrate this one.
  • This yields \( du = \frac{1}{2\sqrt{x}} \, dx \) and \( v = -e^{-x} \).
Now, when you substitute into the formula, you get:
  • \[ \int (1+\sqrt{x})e^{-x}dx = -(1+\sqrt{x})e^{-x} - \int -\frac{e^{-x}}{2\sqrt{x}} \, dx \]
Remember, the key is identifying \( u \) and \( dv \) properly and then carefully applying the formula.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) bridges the concepts of differentiation and integration, emphasizing their inverse relationship. Here's a simple way to understand it:
  • For a continuous function \( f \), it states that if \( F \) is an antiderivative of \( f \), the definite integral of \( f \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).
In this exercise, the function to be integrated is \( (1+\sqrt{x})e^{-x} \), and we found its antiderivative as part of solving the problem.
  • We then applied the FTC by evaluating the antiderivative at the bounds, from 3 to 3.
  • Since the bounds were the same, the integral's value was 0, because \( F(3) - F(3) = 0 \).
This is a crucial point: If the limits of integration are identical, then the definite integral is always zero.
Substitution Method
The Substitution Method, or change of variables, simplifies the integration of complex expressions by substituting a single variable, usually denoted as \( t \) or \( u \), for a part of the integrand.
  • In the integration by parts solution, after initial operations, we faced the integral \( \int -\frac{e^{-x}}{2\sqrt{x}} \, dx \).
  • To simplify it, we used substitution: set \( t = \sqrt{x} \), implying \( x = t^2 \), thus \( dx = 2t \, dt \).
  • This makes the integral easier: \( \int e^{-t^2} \, dt \).
It's essential to recognize that some integrals, like \( \int e^{-t^2} \, dt \), do not have standard elementary antiderivatives, so we often leave them unevaluated in practical settings.
Substitution is a vital tool for transforming difficult integrals into manageable ones, ensuring more straightforward computation.

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