Question

Prove that (i) \(0<\int_{0}^{\pi / 2} \sin ^{n+1} x d x<\int_{0}^{\pi / 2} \sin ^{2} x d x, n>1\) (ii) \(1<\int_{0}^{\pi / 2} \sqrt{\sin x} \mathrm{~d} \mathrm{x}<\sqrt{\frac{\pi}{2}}\) (iii) \(\mathrm{e}^{-\frac{1}{4}}<\int_{0}^{1} \mathrm{e}^{\mathrm{x}^{2}-\mathrm{x}} \mathrm{dx}<1\) (iv) \(-\frac{1}{2} \leq \int_{0}^{1} \frac{x^{3} \cos x}{2+x^{2}} d x<\frac{1}{2}\).

Short answer

#Problem# Determine whether the following inequalities hold for definite integrals: (i) \(0<\int_{0}^{\pi / 2} \sin ^{n+1} x d x<\int_{0}^{\pi / 2} \sin ^{2} x d x, n>1\) (ii) \(1<\int_{0}^{\pi / 2} \sqrt{\sin x} ~d x<\sqrt{\frac{\pi}{2}}\) (iii) \(e^{-\frac{1}{4}}<\int_{0}^{1} e^{x^{2}-x}~dx<1\) (iv) \(-\frac{1}{2} \leq \int_{0}^{1} \frac{x^{3} \cos x}{2+x^{2}} d x<\frac{1}{2}\) #Solution# (i) Yes, the inequality holds because \(0<\sin^{n+1}x<\sin^2x\) for \(n>1\) and \(x\in[0,\pi / 2]\), so we can apply the comparison theorem for integration. (ii) Yes, the inequality holds because \(1<\sin\frac{x}{\sqrt{2}}<\sqrt{\sin x}\) for \(x\in[0,\pi / 2]\), so we can apply the comparison theorem for integration. (iii) Yes, the inequality holds because \(e^{-\frac{1}{4}}<e^{x^2-x}<1\) for \(x\in[0,1]\), so we can apply the comparison theorem for integration. (iv) Yes, the inequality holds because \(-\frac{x^3}{2+x^2}\leq\frac{x^3\cos x}{2+x^2}\leq\frac{x^3}{2+x^2}\) for \(x\in[0,1]\) and after applying the comparison theorem for integration, we obtain \(-\frac{1}{2}\leq\int_{0}^{1}\frac{x^{3}\cos x}{2+x^{2}} dx\leq\frac{1}{2}\).

Step-by-step solution

Compare the integrands

For \(n>1\), we know that \(\sin^{n+1}x < \sin^2x\). Also, \(\sin^{n+1}x > 0\). Thus, we can say that \(0<\sin^{n+1}x<\sin^2x\) for \(x\in[0,\pi / 2]\).

Apply the comparison theorem

Since \(0<\sin^{n+1}x<\sin^2x\) for \(x\in[0,\pi / 2]\), we can apply the comparison theorem for integration and get \(0<\int_{0}^{\pi / 2} \sin^{n+1} x dx<\int_{0}^{\pi / 2} \sin^{2} x dx\). (ii) \(1<\int_{0}^{\pi / 2} \sqrt{\sin x} ~d x<\sqrt{\frac{\pi}{2}}\)

Compare the integrands

Since \(\sqrt{x}>\frac{x}{\sqrt{2}}\) for \(x\in[0,1]\), we can say that \(0<\sin\frac{x}{\sqrt{2}}<\sqrt{\sin x}\) for \(x\in[0,1]\).

Apply the comparison theorem

Since \(0<\sin\frac{x}{\sqrt{2}}<\sqrt{\sin x}\) for \(x\in[0,\pi / 2]\), we can apply the comparison theorem for integration and get \(1<\int_{0}^{\pi / 2} \sqrt{\sin x} dx<\sqrt{\frac{\pi}{2}}\). (iii) \(e^{-\frac{1}{4}}<\int_{0}^{1} e^{x^{2}-x}~dx<1\)

Compare the integrands

Since \(e^{x^2-x}\) increases with \(x\) and \(x\in[0,1]\), we can say that \(e^{-\frac{1}{4}}<e^{x^2-x}<1\) for \(x\in[0,1]\).

Apply the comparison theorem

Since \(e^{-\frac{1}{4}}<e^{x^2-x}<1\) for \(x\in[0,1]\), we can apply the comparison theorem for integration and get \(e^{-\frac{1}{4}}<\int_{0}^{1} e^{x^{2}-x} dx<1\). (iv) \(-\frac{1}{2} \leq \int_{0}^{1} \frac{x^{3} \cos x}{2+x^{2}} d x<\frac{1}{2}\)

Compare the integrands

Since \(-1\leq \cos x\leq 1\) and \(\frac{x^3}{2+x^2}\) is always positive in the range \(x\in[0,1]\), we can say that \(-\frac{x^3}{2+x^2}\leq\frac{x^3\cos x}{2+x^2}\leq\frac{x^3}{2+x^2}\) for \(x\in[0,1]\).

Integrate the upper and lower bounds

We know that \(-\frac{x^3}{2+x^2}\leq\frac{x^3\cos x}{2+x^2}\leq\frac{x^3}{2+x^2}\). Now we will integrate these inequalities with respect to \(x\) over the interval \([0,1]\). Applying the comparison theorem for integration, we will get the definite integral. Solve these integrals separately and get \(-\frac{1}{2}\leq\int_{0}^{1}\frac{x^{3}\cos x}{2+x^{2}} dx\leq\frac{1}{2}\).

Key concepts

Comparison Theorem for Integration

The Comparison Theorem for Integration is a powerful tool that aids in evaluating the bounds of a definite integral when the integral itself is difficult to compute directly. It states that if we have two continuous functions, say, \f\(f(x)\f\) and \f\(g(x)\f\), on an interval [a, b] and \f\(0 \fleq f(x) \fleq g(x)\f\) for all \f\(x \fleqin [a, b]\f\), then their definite integrals also maintain that order: \f\(0 \fleq \fint_{a}^{b} f(x) dx \fleq \fint_{a}^{b} g(x) dx\f\).

This theorem can be particularly useful when confronted with challenging integrals. By comparing a tough integrand with a simpler one that we can integrate, we can establish lower and upper bounds for our integral, thus obtaining an estimate for its value.

For example, if we know that every function \f\(sin^{n+1}x\f\) lies between 0 and \f\(sin^2x\f\) within the interval [0, \f\(\fpi / 2]\f\), as it happens with the integral in part (i) of our exercise, we can use the comparison theorem to conclude that the integral of \f\(sin^{n+1}x\f\) must also lie between the integrals of these two bounds.

Definite Integrals

Definite integrals provide the total accumulation of a quantity, which could represent area, volume, or other concepts, between two points along the x-axis. Defined by two limits, an upper and a lower, definite integrals represent the sum of infinitesimal products over a certain interval.

Mathematically, a definite integral is expressed as \f\(\int_{a}^{b}f(x)dx\f\), where \f\(a\f\) and \f\(b\f\) are the lower and upper limits, respectively, and \f\(f(x)\f\) is the integrand. Part (ii) of our exercise illustrates a scenario where the value of a definite integral, \f\(\int_{0}^{\fpi / 2} \fendsqrt{\sin x} dx\f\), is sought within known boundaries, giving us a range rather than an exact figure. Definite integrals' capability to encapsulate continuous phenomena in this bounded manner makes them an indispensable tool in calculus and applied mathematics.

Inequalities in Integration

Inequalities in integration play a crucial role when estimating or bounding the values of integrals. They come into effect under the premise that if one function is greater than another in an interval, then the integral of the first will also be greater than the integral of the second, across the same interval.

Mathematically, if \f\(f(x) > g(x)\f\) for all \f\(x\f\) in the interval [a, b], then \f\(\int_{a}^{b}f(x)dx > \int_{a}^{b}g(x)dx\f\). However, this only holds if the functions involved are integrable over the interval and do not change signs. In our exercises, this principle is used to establish the inequalities presented in parts (iii) and (iv), where the integrals of exponential and trigonometric functions are bounded by simpler expressions, leveraging the properties of these functions to derive conclusions about the ranges of these integrals.

Integration Techniques

Integration techniques are various methods used to calculate integrals, especially when the function involved is complex. These include methods like substitution, integration by parts, partial fraction decomposition, and trigonometric substitution, among others. Each technique has its use-case scenarios, with substitution being one of the most commonly employed for simplifying the integrand to a more recognizable form.

For instance, in more challenging problems, you may integrate a function by recognizing its derivative within the integral and substituting to make the problem simpler, as you would with a definite integral involving a complex exponent.

Understanding when and how to apply different integration techniques is crucial for solving a wide range of problems. In the context of our exercise, such techniques might be helpful in finding the definite integrals that bound the inequalities, had the comparison theorem not provided a simpler pathway to the solution.