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Chapter 8: Problem 79

Evaluate the integrals. \begin{equation} \text { a. }\int_{0}^{1} \frac{d t}{\sqrt{t}(1+t)} \quad \text { b. } \int_{0}^{\infty} \frac{d t}{\sqrt{t}(1+t)} \end{equation}

Short Answer

Expert verified
a. \(\frac{\pi}{2}\), b. \(\pi\).

Step by step solution

01

Solve Integral (a)

The integral is given as \( \int_{0}^{1} \frac{d t}{\sqrt{t}(1+t)} \). We'll use substitution to solve it. Let \( u = \sqrt{t} \), so \( t = u^2 \) and \( dt = 2u \, du \). The limits of integration also change: when \( t = 0 \), \( u = 0 \); and when \( t = 1 \), \( u = 1 \).Substitute into the integral:\[ \int_0^1 \frac{2u}{u(1+u^2)} \, du = \int_0^1 \frac{2}{1+u^2} \, du. \] This is a standard integral, which simplifies to:\[ 2 \tan^{-1}(u) \Big|_0^1. \] Calculate the definite integral:\[ 2 \left( \tan^{-1}(1) - \tan^{-1}(0) \right) = 2 \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{2}. \]
02

Solve Integral (b)

The integral is given as \( \int_{0}^{\infty} \frac{d t}{\sqrt{t}(1+t)} \) which is improper. We employ a similar substitution as before: let \( u = \sqrt{t} \), so \( t = u^2 \) and \( dt = 2u \, du \). The limits change to infinity: \( u = 0 \) when \( t = 0 \) and \( u = \infty \) when \( t = \infty \).The integral becomes:\[ \int_0^\infty \frac{2u}{u(1+u^2)} \, du = \int_0^\infty \frac{2}{1+u^2} \, du. \] This is a standard form which evaluates to:\[ 2 \tan^{-1}(u) \Big|_0^\infty. \]Calculating this:\[ 2 \left( \frac{\pi}{2} - 0 \right) = \pi. \] Thus, the value of the integral is \( \pi \).

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Key Concepts

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

Definite Integrals
Definite integrals are a key concept in calculus, dealing with the computation of the area under a curve between two specific limits. In this exercise, we first addressed a definite integral, which means the integral has specified upper and lower limits. For example, the integral \[\int_0^1 \frac{d t}{\sqrt{t}(1+t)}\] from 0 to 1, is a definite integral. Here's what you need to know about them:
  • The definite integral computes the net area between the function and the x-axis from the lower to the upper limit.
  • It is expressed as \(\int_a^b f(x) \, dx\), where \(a\) and \(b\) are the limits, and \(f(x)\) is the function.
  • Fundamentally, they are used to calculate quantities like area, volume, and other related measures.
The fundamental theorem of calculus helps to evaluate these integrals efficiently, linking the concept of derivative with the integral.
Substitution Method
The substitution method is a powerful technique used to simplify integrals, which are otherwise complex to solve. It involves changing variables to make the integral easier to evaluate. Let鈥檚 dive into how it works through an example from the solution given.
To solve the integral \(\int \frac{d t}{\sqrt{t}(1+t)}\), the substitution \(u = \sqrt{t}\) is used. This transforms the integral into a more manageable form:
  • Express the variable to be substituted in terms of \(u\), here \(t = u^2\).
  • Convert \(dt\) as well, yielding \(dt = 2u \, du\).
  • Substitute these into the integral, changing it to \(\int \frac{2}{1+u^2} \, du\).
Then calculate the integral using basic integration rules or known results, such as the integral of \(\frac{1}{1+u^2}\) being \(\tan^{-1}(u)\). Substitution is particularly useful when dealing with trigonometric and rational functions.
Improper Integrals
Improper integrals extend the concept of integration to unbounded intervals or singular points within the range of integration. This occurs when either the integration limits are infinite, or the function approaches infinity within the integration limits.
In the exercise provided, the integral \(\int_0^\infty \frac{d t}{\sqrt{t}(1+t)}\) is an improper integral due to the upper limit of infinity.
  • To handle these, often we need to evaluate a limit as one of the bounds approaches infinity, or as the function approaches a point of discontinuity.
  • This involves calculating the limit of the definite integral as it stretches towards infinity.
  • In our example, after the substitution method turned the integral into a standard form, we calculated the limit as the upper bound goes to infinity.
These integrals are essential in fields requiring total sums over indefinite successions, like physics and probability.

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