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Chapter 7: Problem 8

Calculate the following iterated integrals. $$ \int_{0}^{1}\left(\int_{0}^{x} e^{x+y} d y\right) d x $$

Short Answer

Expert verified
\( \frac{1}{2} e^2 - e + \frac{1}{2} \)

Step by step solution

01

Understand the Inner Integral

Identify the inner integral \(\int_{0}^{x} e^{x+y} dy\). Treat \(x\) as a constant while integrating with respect to \(y\).
02

Integrate the Inner Integral

Integrate \(e^{x+y}\) with respect to \( y \). Since \(x\) is constant, we have \(\int_{0}^{x} e^{x+y} dy = e^x \int_{0}^{x} e^{y} dy = e^x \big|_{0}^{x} e^y \).
03

Evaluate the Inner Integral

Evaluate the result of the inner integral: \(e^x \big|_{0}^{x} e^{y} = e^x (e^x - e^0) = e^x (e^x - 1)\). This simplifies to \(e^{2x} - e^x\).
04

Understand the Outer Integral

The outer integral now becomes \(\int_{0}^{1} (e^{2x} - e^x) dx \).
05

Integrate the Outer Integral

Integrate \(e^{2x} - e^x\) with respect to \(x\). This gives \(\int_{0}^{1} e^{2x} dx - \int_{0}^{1} e^x dx \).
06

Evaluate the Outer Integral

Evaluate \( \int_{0}^{1} e^{2x} dx \) and \( \int_{0}^{1} e^x dx \): \( \int_{0}^{1} e^{2x} dx = \frac{1}{2} e^{2x} \bigg|_{0}^{1} \) giving \( \frac{1}{2} (e^2 - 1) \), and \( \int_{0}^{1} e^x dx = e^x \bigg|_{0}^{1} = e - 1 \).
07

Combine and Simplify

Combine the results: \(\frac{1}{2} (e^2 - 1) - (e - 1) = \frac{1}{2}e^2 - \frac{1}{2} - e + 1 = \frac{1}{2} e^2 - e + \frac{1}{2} \).

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Key Concepts

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

Integration by Parts
Integration by parts is a powerful technique used in calculus to integrate products of functions. It's based on the product rule for differentiation. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du\]\. To break this down:
  • Choose parts of the integrand as \(u\) and \(dv\).
  • Differentiate \(u\) to find \(du\).
  • Integrate \(dv\) to find \(v\).
  • Plug the parts into the formula.
  • Simplify and integrate any remaining terms.
This method is especially useful when the integral of a product of functions isn't straightforward. In our problem, integration by parts is used in the outer integral \int_{0}^{1} (e^{2x} - e^x) dx. Each function, \(e^{2x}\) and \(e^x\), needs to be integrated separately.
Definite Integrals
Definite integrals represent the area under a curve within a specified range. For an integral \[ \int_{a}^{b} f(x) \, dx \], \(a\) and \(b\) are the limits of integration. The process involves:
  • Finding the antiderivative of \(f(x)\), denoted as \[ F(x) \].
  • Evaluating \(F(x)\) at the upper limit \(b\) and the lower limit \(a\).
  • Subtracting the value at \(a\) from the value at \(b\), written as \[F(b) - F(a). \]
In our problem, we calculated \int_{0}^{x} e^{x+y} dy \ as part of finding the inner integral. When evaluating a definite integral, ensure to carefully follow the limits, and correctly substitute into the antiderivative.
Exponential Functions
Exponential functions have the form \[ e^x \]. They are unique in calculus due to their property: the derivative and the integral of \[ e^x \] are both \[ e^x \]. This characteristic simplifies many integration problems.

However, complications can arise when the exponent is more complex, such as \[ e^{2x} \]. In these cases:
  • Apply the chain rule for differentiation.
  • Consider substitution methods for integration.
In our problem, integrating \[ e^{2x} \] required using the substitution method, where we accounted for a factor outside the exponential function. Being familiar with these rules greatly aids in solving integrals involving exponential functions.

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Most popular questions from this chapter

Find all points \((x, y)\) where \(f(x, y)\) has a possible relative maximum or minimum. $$ f(x, y)=x^{3}+y^{2}-3 x+6 y $$

Compute \(\frac{\partial^{2} f}{\partial y^{2}}\), where \(f(x, y)=60 x^{3 / 4} y^{1 / 4}\), a production function (where \(y\) is units of capital). Explain why \(\frac{\partial^{2} f}{\partial y^{2}}\) is always negative.

Both first partial derivatives of the function \(f(x, y)\) are zero at the given points. Use the second-derivative test to determine the nature of \(f(x, y)\) at each of these points. If the second-derivative test is inconclusive, so state. $$ f(x, y)=6 x y^{2}-2 x^{3}-3 y^{4} ;(0,0),(1,1),(1,-1) $$

Find all points \((x, y)\) where \(f(x, y)\) has a possible relative maximum or minimum. Then, use the second-derivative test to determine, if possible, the nature of \(f(x, y)\) at each of these points. If the second-derivative test is inconclusive, so state. $$ f(x, y)=-2 x^{2}+2 x y-y^{2}+4 x-6 y+5 $$

Calculate the following iterated integrals. $$ \int_{-1}^{1}\left(\int_{-1}^{1} x y d x\right) d y $$
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