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Chapter 4: Problem 51

For the integral \(I=\int_{0}^{10} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x,\) use a substitution to show that \(I=\int_{0}^{10} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x .\) Use these two representations of 1 to evaluate \(I\)

Short Answer

Expert verified
The result of the integral \(I\) is 5.

Step by step solution

01

Transformation

The idea is to make a transformation \( u = 10 - x \) which simplifies the integral. To proceed with this, we differentiate \( u \) with respect to \(x\), i.e., \( du = -dx \). Then, rearrange to find \( dx \) in terms of \( du \), i.e., \( dx = -du \).It's also necessary to change the limits of integration. When \( x = 0 \), \( u = 10 \), and when \( x = 10 \), \( u = 0 \).
02

Substituting in the integral

Substitute \( u \), \( u \), and the limits in the original integral, obtaining \( \int_{10}^{0} \frac{\sqrt{10 - u}}{\sqrt{10 - u} + \sqrt{u}} \cdot - du \). The negative sign can be used to switch the limits of integration, which makes it \( \int_{0}^{10} \frac{\sqrt{10 - u}}{\sqrt{10 - u} + \sqrt{u}} du \), similar to the required integral in the exercise.
03

Final verification

We have two integrals \(I=\int_{0}^{10} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} dx \) and \(I=\int_{0}^{10} \frac{\sqrt{10 - x}}{\sqrt{x}+\sqrt{10 - x}} dx \). Adding these two integrals we get \(2I = \int_{0}^{10} \frac{\sqrt{x}+\sqrt{10 - x}}{\sqrt{x}+\sqrt{10 - x}} dx = \int_{0}^{10} dx = 10\). Hence, \(I = 5 \).

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Key Concepts

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

Substitution Method
In integral calculus, the substitution method is a powerful technique used to simplify integration. It is akin to the reverse of the chain rule used in differentiation. The main idea is to substitute a part of the original integral with a new variable, which simplifies the expression and makes integration more straightforward.
For example, consider the transformation used in the provided exercise: you use the substitution \( u = 10 - x \). This transforms the original complex integral into a simpler form. When performing a substitution:
  • Identify a portion of the integral that can be replaced (typically, inside a composite function).
  • Define a new variable \( u \) and express the differential \( dx \) in terms of \( du \).
  • Replace the original limits of integration with new ones that correspond to the \( u \)-variable.
This technique allows for simplification by often removing square roots, products, or other complex terms. Always remember to revert back to the original variable if the integral is indefinite. But for definite integrals, after evaluating, the substitution directly gives the result in terms of the new limits.
Definite Integrals
Definite integrals in calculus represent the accumulation of quantities and can be used to find areas under curves on a graph. They are denoted by the integral sign with upper and lower limits, \( \int_{a}^{b} f(x)\, dx \). These limits (\( a \) and \( b \)) serve as the boundaries for integration.
In our exercise, the definite integral \( \int_{0}^{10} \) has boundary values 0 and 10. These represent the start and end points on the \( x \) axis. When computing a definite integral:
  • Evaluate the antiderivative at the upper limit \( b \).
  • Subtract from it the antiderivative evaluated at the lower limit \( a \).
The result is a finite number representing the total accumulation between \( a \) and \( b \). Changing these limits, as done in the substitution method, does not affect the overall outcome because integration is path-independent when applied correctly. The symmetry in the integral here demonstrates how clever substitution can reveal hidden relationships that simplify the problem.
Limits of Integration
The limits of integration determine the interval over which the function is being integrated. They play a crucial role in defining what portions of the continuous function contribute to the final integral value. In our example, the limits are changed through substitution from \( x \) to \( u \), affecting how the integral is approached and solved.
During substitution, you must adjust these limits:
  • Calculate new lower and upper limits by substituting the original limits into the expression for \( u \).
  • After substitution, solve the integral from these new limits to ensure consistency in the integral's value.
For example, what started as \( \int_{0}^{10} f(x)\, dx \) changed into \( \int_{10}^{0} g(u)\, du \). However, applying the negative sign straightens these limits to \( \int_{0}^{10} g(u)\, du \), keeping the results equivalent. Proper management of these limits ensures that the substitution method and the evaluation of definite integrals yield correct solutions.

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Most popular questions from this chapter

The number of items that consumers are willing to buy depends on the price of the item. Let \(p=D(q)\) represent the price (in dollars) at which \(q\) items can be sold. The integral \(\int_{0}^{Q} D(q) d q\) is interpreted as the total number of dollars that consumers would be willing to spend on \(Q\) items. If the price is fixed at \(P=D(Q)\) dollars, then the actual amount of money spent is \(P Q .\) The consumer surplus is defined by \(C S=\int_{0}^{Q} D(q) d q-P Q .\) Compute the consumer surplus for \(D(q)=150-2 q-3 q^{2}\) at \(Q=4\) and at \(Q=6 .\) What does the difference in \(C S\) values tell you about how many items to produce?

Involve the just-in-time inventory discussed in the chapter introduction. For a business using just-in-time inventory, a delivery of \(Q\) items arrives just as the last item is shipped out. Suppose that items are shipped out at the constant rate of \(r\) items per day. If a delivery arrives at time \(0,\) show that \(f(t)=Q-r t\) gives the number of items in inventory for \(0 \leq t \leq \frac{Q}{r} .\) Find the average value of \(f\) on the interval \(\left[0, \frac{Q}{r}\right]\).

Let \(f\) be a continuous function on the interval \([0,1],\) and \(\mathrm{de}-\) fine \(g_{n}(x)=f\left(x^{n}\right)\) for \(n=1,2\) and so on. For a given \(x\) with \(0 \leq x \leq 1,\) find \(\lim _{n \rightarrow \infty} g_{n}(x) .\) Then, find \(\lim _{n \rightarrow \infty} \int_{0}^{1} g_{n}(x) d x\)

Evaluate the integral. $$\int_{e}^{e^{2}} \frac{1}{x \ln x} d x$$

Starting with \(\quad e^{x}=\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}, \quad\) show \(\quad\) that \(\ln x=\lim _{n \rightarrow \infty}\left[n\left(x^{1 / n}-1\right)\right] .\) Assume that if \(\lim _{n \rightarrow \infty} x_{n}=x,\) then \(\lim _{n \rightarrow \infty}\left[n\left(x_{n}^{1 / n}-1\right)\right]=\lim _{n \rightarrow \infty}\left[n\left(x^{1 / n}-1\right)\right].\)
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