Question

Prove that (i) \(0<\int_{0}^{2} \frac{x \mathrm{dx}}{16+x^{3}}<\frac{1}{6}\) (ii) \(\int_{0}^{1} \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^{4}}} \geq \frac{\pi}{4}\) (iv) \(\int_{0}^{1} \frac{d x}{4+x^{3}}>\ln \frac{5}{4}\) (iii) \(\int_{1}^{100} \mathrm{e}^{-\mathrm{x}} \sin ^{2} \mathrm{x} \mathrm{dx}<1\)

Short answer

#Answer#: (i) Following the steps in the given solution, we found that \(0<\int_{0}^{2} \frac{x \mathrm{dx}}{16+x^{3}}<\frac{1}{6}\) by showing the integral is positive and finding an upper bound. (ii) To prove that \(\int_{0}^{1} \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^{4}}} \geq \frac{\pi}{4}\), we found a lower bound, computed the integral, and made the comparison to \(\frac{\pi}{4}\). (iii) We proved that \(\int_{1}^{100} \mathrm{e}^{-\mathrm{x}} \sin ^{2} \mathrm{x} \mathrm{dx}<1\) by establishing an upper bound and comparing the integral of the upper bound to 1. (iv) Lastly, we showed that \(\int_{0}^{1} \frac{d x}{4+x^{3}}> \ln \frac{5}{4}\) by finding a lower bound, computing the integral, and comparing it to \(\ln\frac{5}{4}\).

Step-by-step solution

Show the integral is positive

Let's consider the function \(\frac{x}{16+x^{3}}\). Since \(x\ge0\), and \(16+x^{3}>0\) for \(x\in[0,2]\), the function is positive on the given interval. Because the function is positive, the definite integral over this interval is positive as well.

Find an upper bound

We need to find an upper bound \(f(x)\) for the function \(g(x)=\frac{x}{16+x^{3}}\) on the interval \([0,2]\), such that \(\int_{0}^{2} f(x) \mathrm{dx} = \frac{1}{6}\). To achieve this, we notice that \(16+x^3 > 8x\) for \(x\in[0,2]\), leading us to the inequality \(\frac{x}{16+x^3} < \frac{x}{8x} = \frac{1}{8}\). Therefore,\(f(x)=\frac{1}{8}\) is an upper bound of \(g(x)\) on the given interval. Now let's compute the integral of \(f(x)\): \(\int_{0}^{2} \frac{1}{8} \mathrm{dx} = \frac{1}{8}(2) = \frac{1}{4} > \frac{1}{6}\) So, we have demonstrated that \(0<\int_{0}^{2} \frac{x \mathrm{dx}}{16+x^{3}}<\frac{1}{6}\). (ii) Proving \(\int_{0}^{1} \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^{4}}} \geq \frac{\pi}{4}\)

Find a lower bound

We need to find a lower bound \(f(x)\) for the function \(g(x)=\frac{1}{\sqrt{1+x^{4}}}\) on the interval \([0,1]\). To achieve this, we notice that \(x^4 \le x^2\) for \(x\in[0,1]\). Therefore, \(1+x^4 \ge 1+x^2\), leading us to the inequality \(\frac{1}{\sqrt{1+x^4}} \ge \frac{1}{\sqrt{1+x^2}}\). Now consider the lower bound \(f(x) = \frac{1}{\sqrt{1+x^2}}\).

Compute the integral and compare

Compute the integral of the lower bound \(f(x)\) over the interval \([0,1]\) and compare it to \(\frac{\pi}{4}\). \(\int_{0}^{1} \frac{1}{\sqrt{1+x^2}} \mathrm{dx} = [\arcsin(x)]_{0}^{1} = \arcsin(1) - \arcsin(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}\) Since \(\frac{\pi}{2} > \frac{\pi}{4}\), we have shown that \(\int_{0}^{1} \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^{4}}} \geq \frac{\pi}{4}\). (iii) Proving \(\int_{1}^{100} \mathrm{e}^{-\mathrm{x}} \sin ^{2} \mathrm{x} \mathrm{dx}<1\)

Establish an upper bound

Notice that \(0 \le \sin^2(x) \le 1\) for all \(x\in[1,100]\). So, we can establish an upper bound as follows: \(\mathrm{e}^{-x} \sin^2(x) \le \mathrm{e}^{-x}\).

Compute the integral of the upper bound and compare

Now, compute the integral of the upper bound \(\mathrm{e}^{-x}\) over the interval \([1, 100]\) and compare it to \(1\): \(\int_{1}^{100} \mathrm{e}^{-x} \mathrm{dx} = [-\mathrm{e}^{-x}]_{1}^{100} = -\mathrm{e}^{-100} + \mathrm{e}^{-1} \approx 0.632 < 1\) Since the integral of the upper bound is less than \(1\), the integral of the original function is also less than \(1\). Therefore, \(\int_{1}^{100} \mathrm{e}^{-\mathrm{x}} \sin ^{2} \mathrm{x} \mathrm{dx}<1\). (iv) Proving \(\int_{0}^{1} \frac{d x}{4+x^{3}}> \ln \frac{5}{4}\)

Find a lower bound

We need to find a lower bound \(f(x)\) for the function \(g(x)=\frac{1}{4+x^{3}}\) on the interval \([0,1]\). Since \(x^3 \le x\) for \(x\in[0,1]\), we have \(4+x^3 \ge 4+x\), leading us to the inequality \(\frac{1}{4+x^3} \ge \frac{1}{4+x}\). Now consider the lower bound \(f(x) = \frac{1}{4+x}\).

Compute the integral and compare

Compute the integral of the lower bound \(f(x)\) over the interval \([0,1]\) and compare it to \(\ln \frac{5}{4}\). \(\int_{0}^{1} \frac{1}{4+x} \mathrm{dx} = [\ln(4+x)]_{0}^{1} = \ln(5) - \ln(4) = \ln\frac{5}{4}\) Since \(\int_{0}^{1} \frac{d x}{4+x^{3}} > \int_{0}^{1} \frac{1}{4+x} \mathrm{dx} = \ln\frac{5}{4}\), we have shown that \(\int_{0}^{1} \frac{d x}{4+x^{3}}> \ln \frac{5}{4}\).

Key concepts

Integral Inequalities

Integral inequalities are a powerful tool in calculus, particularly in evaluating definite integrals without finding the explicit integral function. The essence of this concept is to establish bounds for the definite integral of a function over an interval. To use integral inequalities effectively, one should identify an easily integrable function that bounds the given function either from above or below on the specified interval.

As illustrated in the IIT JEE problems, a common strategy is to manipulate the integrand to find a new integrand that serves as an upper or lower bound, which is more straightforward to integrate. For instance, using the inequality that if a function is positive over a range, its integral over that range is also positive, helps in establishing one side of the inequality. Furthermore, recognising that the integrals of these bounding functions overestimate or underestimate the integral of the original function solidifies the inequality. This step-by-step reasoning not only meets the goal of homework exercises but also strengthens conceptual understanding, which is essential for tackling a variety of problems in the IIT JEE Mathematics syllabus.

Integral Calculus

Integral calculus, one of the two principal branches of calculus, revolves around the accumulation of quantities and the areas under and between curves. In the context of IIT JEE, it encompasses concepts such as definite and indefinite integrals, fundamental theorems of calculus, and application of integrals to physical problems.

Solutions to integral calculus problems on the IIT JEE often require a blend of techniques—such as substitution, partial fractions, and integration by parts—as well as conceptual understanding of bounding functions and the behavior of functions on intervals, as in integral inequalities. Emphasis on clear, step-by-step solutions can help students grasp the intricate processes, especially when working with inequalities in integrals and estimating definite integrals. Simplification and logical breakdown are pivotal in transforming a complex problem into manageable steps.

IIT JEE Mathematics

The IIT Joint Entrance Examination (IIT JEE) is one of the most challenging engineering entrance exams in India, and mathematics is a critical component of it. IIT JEE Mathematics covers a wide range of topics, including algebra, calculus, coordinate geometry, trigonometry, vector algebra, and probability, among others.

Mastering definite integrals, as seen in the example problems, involves not only computational skills but also analytical thinking to work through complex problems. Such problem-solving capabilities are nurtured through regular practice and exposure to varied types of questions. One improvement advice for students preparing for this section is to delve deeply into core concepts, real-world applications, and practice exercises with incremental difficulty levels to build a profound understanding of the subject.

Integral Bounds

Integral bounds refer to the upper and lower limits within which the value of a definite integral exists. Determining accurate integral bounds is crucial for managing computational complexity and is particularly useful in establishing inequalities. When the exact value of an integral is arduous to compute or unnecessary for the particular context, bounds provide a simplification that streamlines the solution process.

For example, in the provided IIT JEE problems, establishing that the integral of a function is greater than one value and less than another is facilitated by integral bounds. Ensuring that the bounds are appropriately tight, meaning close enough to the actual value of the integral, is key to the usefulness of this approach. This is where knowledge of inequities and properties of functions, such as their monotonous nature or positivity, play a vital role and are often tested in the IIT JEE Mathematics curriculum.