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Chapter 14: Problem 24

Evaluate the following integrals as they are written. $$\int_{0}^{\ln 2} \int_{e^{x}}^{2} d y d x$$

Short Answer

Expert verified
Question: Evaluate the double integral given by $$\int_{0}^{\ln2}\int_{e^{x}}^{2} dy dx$$ Answer: The value of the double integral is $$2\ln 2 - 3$$.

Step by step solution

01

Integrate with respect to the inner variable, y

Since there is no function inside the integral, we can just integrate 1 with respect to y. The integral of 1 with respect to y is y: $$\int \int 1 dy dx = \int (y) dx$$ Now we need to find the integral between the given limits for y: $$\int_{0}^{\ln 2} (2 - e^{x}) d x$$
02

Integrate with respect to the outer variable, x

Now, we find the integral of the function (2 - e^x) with respect to x: $$\int (2 - e^{x}) d x = 2x - e^{x} + C$$
03

Find the definite integral for the given limits

Now, we need to find the value of the integral between the limits 0 and ln(2) for x: $$\int_{0}^{\ln 2} (2 - e^{x}) d x = [2x - e^{x}]_{0}^{\ln 2}$$ Now plug in the upper limit, ln(2), and subtract the result when we plug in the lower limit, 0: $$\left(2(\ln 2) - e^{\ln 2}\right) - \left(2(0) - e^{0}\right) = 2\ln 2 - 2 - 1$$
04

Simplify the final answer

Finally, let's simplify the expression: $$2\ln 2 - 2 - 1 = 2\ln 2 - 3$$ So, the final answer is: $$\int_{0}^{\ln2}\int_{e^{x}}^{2} dy dx = 2\ln 2 - 3$$

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Key Concepts

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

Understanding Integral Calculus
Integral calculus is a branch of calculus concerned with the theory of integration and its applications. It revolves around the idea of finding the area under a curve. This area represents the integral of a function over a given interval.

In a double integral, like the one in our exercise, we handle functions of two variables. The notation \(\int_{a}^{b}\int_{c}^{d} f(y, x) \, dy \, dx\)is typically used, where we first integrate with respect to \(y\), and then \(x\).
  • This process can be visualized as accumulating small strips in the \(y\)-direction, and stacking them along the \(x\)-direction.
  • The double integral essentially helps find volumes under surfaces defined by \(f(y, x)\).
  • It's crucial to evaluate these integrals in the correct order unless the limits and function allow for a change of order.
You will see that these integrations occur step-by-step as you go from inside out, which simplifies solving complex problems.
Definite Integrals Explained
Definite integrals are a type of integral with set limits, defining the upper and lower bounds of the area we wish to compute. When you see notations such as \(\int_{a}^{b} f(x) \, dx\), it represents the area from \(a\) to \(b\) under the curve of \(f(x)\).

Compared to indefinite integrals, definite integrals give a numerical value rather than a function. In our exercise:
  • The outer integral limits, from \(0\) to \(\ln 2\), represent the bounds for \(x\).
  • The inner limits, from \(e^x\) to \(2\), define the bounds for \(y\).
The importance here is understanding that the definite integral's result tells us the exact accumulated area or volume within those bounds. This is especially useful in applications such as physics and engineering, where we calculate quantities over a specific interval.
Step-by-Step Evaluation of Integrals
Evaluating an integral involves step-by-step processes to find the accumulated sum described by the integral expression. Let's break down the steps further:

First, when you encounter a double integral:
  • Start by integrating with respect to the innermost variable (here it's \(y\)). This simplification usually turns the inner problem into a basic one-variable integral.
  • The forms of upper and lower limits often depend on the other variable, making your \(dy\)integral more like a parameter problem.
In the second step, you find:
  • The integral with respect to the outer variable (\(x\) in this example). This usually involves substituting back the results from your first integration.
  • Use the antiderivative to plug in the boundary values. Remember to subtract initial values from final values after substitution.
Completing these steps ensures you have handled all parts of the expression properly. The result is a single number that tells the total sum of the regions evaluated by the original double integral setup.

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Most popular questions from this chapter

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the cardioid \(r=1+\cos \theta\)

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Find the volume of a tetrahedron whose vertices are located at \((0,0,0),(a, 0,0),(0, b, 0),\) and \((0,0, c)\)

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the transformation \(T: x=g(u, v), y=h(u, v)\) is linear in \(u\) and \(v,\) then the Jacobian is a constant. b. The transformation \(x=a u+b v, y=c u+d v\) generally maps triangular regions to triangular regions. c. The transformation \(x=2 v, y=-2 u\) maps circles to circles.

Evaluate the following integrals using a change of variables of your choice. Sketch the original and new regions of integration, \(R\) and \(S\). \(\iint_{R} e^{x y} d A,\) where \(R\) is the region bounded by the hyperbolas \(x y=1\) and \(x y=4,\) and the lines \(y / x=1\) and \(y / x=3\)

Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a solid ellipsoid with axes of length \(2 a, 2 b,\) and \(2 c\).
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