Question

Evaluate the following integrals: (i) \(\int_{-1}^{0} \frac{e^{\frac{1}{x}}}{x^{3}} d x\) (ii) \(\int_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x}} d x\) (iii) \(\int_{3}^{5} \frac{x^{2} d x}{\sqrt{(x-3)(5-x)}}\) (iv) \(\int_{-1}^{1} \frac{d x}{(2-x) \sqrt{1-x^{2}}}\)

Short answer

In summary, the definite integral values for each problem are as follows: 1. The numerical value for the integral using numerical integration methods. 2. \(\frac{\pi}{2}\) 3. \(\frac{392}{3}\) 4. 0

Step-by-step solution

Identify the function and limits

In this problem, the function to be integrated is \(\frac{e^{\frac{1}{x}}}{x^{3}}\), and the limits of integration are -1 and 0.

Perform the integration

In this case, it is quite challenging to find an analytical solution for this integral. Therefore, we recommend using numerical integration methods such as Simpson's rule or the trapezoidal rule, which can be done using a computer program or calculator.

Apply limits of integration

Once we have evaluated the integral using a numerical integration method, we can apply the limits to find the definite integral value. (ii) Integral Problem 2:

Identify the function and limits

In this problem, the function to be integrated is \(\frac{1}{e^{x}+e^{-x}}\), and the limits of integration are \(-\infty\) and \(\infty\).

Perform the integration

This integral can be solved using a substitution. Let \(u = e^x\). Then, \(du = e^x dx\) and \(dx = \frac{du}{e^x}\). We also change the limits of integration; we get \(u = e^{-\infty} = 0\) and \(u = e^{\infty} = \infty\). Now substitute in the integral and we get: $$\int_0^{\infty} \frac{1}{u + \frac{1}{u}} \frac{du}{u}$$ We can simplify this integral as: $$\int_0^{\infty} \frac{1}{1 + \frac{1}{u^2}} du$$ This integral is known as the Integral of the Hyperbolic Tangent function, and it is equal to: $$\frac{\pi}{2}$$

Apply limits of integration

The definite integral value for this problem is \(\frac{\pi}{2}\). (iii) Integral Problem 3:

Identify the function and limits

In this problem, the function to be integrated is \(\frac{x^2}{\sqrt{(x-3)(5-x)}}\), and the limits of integration are 3 and 5.

Perform the integration

For this problem, let \(u = \sqrt{(x-3)(5-x)}\). Then \(du = \frac{(x-3) + (5-x)}{4 \sqrt{(x-3)(5-x)}} dx\). Substituting into the integral: $$\int_3^5 \frac{x^2}{u} \frac{4u}{x-3 + 5-x} du$$ This integral simplifies to: $$4\int_3^5 x^2 du$$ Now we can integrate: $$4\left[\frac{x^3}{3}\right]_3^5$$

Apply limits of integration

Apply the limits and obtain the value for the definite integral: $$4\left[\frac{5^3}{3} - \frac{3^3}{3}\right] = \frac{4}{3} (125 - 27) = \frac{392}{3}$$ (iv) Integral Problem 4:

Identify the function and limits

In this problem, the function to be integrated is \(\frac{1}{(2-x) \sqrt{1-x^2}}\), and the limits of integration are -1 and 1.

Perform the integration

We can use the substitution \(x = \sin u\): \(dx = \cos u du\) and we'll obtain \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{(2-\sin u)\cos u} du\). Now use the substitution \(t = \tan\frac{u}{2}\), with \(dt = \frac{1}{2}(\sec^2\frac{u}{2}) du\). We get: $$\int_{-\infty}^{\infty} \frac{2}{\left(2-\frac{2t}{1+t^2}\right) \left(\frac{1-t^2}{1+t^2}\right)\left(\frac{2}{1+t^2}\right)} dt$$ This integral simplifies to: $$\int_{-\infty}^{\infty} \frac{1}{(1-t)(1+t)} dt$$ Separating into partial fractions, the integral becomes: $$\int_{-\infty}^{\infty} \frac{1}{2}\left(\frac{1}{1 - t} - \frac{1}{1 + t}\right) dt$$ Defining two integrals \(I_1 = \int_{-\infty}^{\infty} \frac{1}{2(1 - t)}dt\) and \(I_2 = \int_{-\infty}^{\infty} \frac{-1}{2(1 + t)}dt\), we can solve both integrals. It turns out that \(I_1\) and \(I_2\) are convergent and both equal to \(\pi\). Thus, the original integral is equal to \(\pi(-1) - \pi = 0\)

Apply limits of integration

As we have evaluated the integral, the definite integral value for this problem is 0.

Key concepts

Numerical Integration Methods

When we're faced with an integral that's difficult or impossible to solve analytically, numerical integration methods come to our rescue. These techniques approximate the integral's value using basic geometric shapes to model the area under a curve. By breaking down an integral into smaller parts, we can sum their areas to approximate the whole.

One popular numerical integration method is Simpson's Rule, which uses parabolic arcs instead of straight lines to approximate the curve. It provides a more accurate result for fewer subdivisions than the Trapezoidal Rule, another common technique that approximates the area using trapezoids. Both require the function to be sampled at various points, and as the number of samples increases, the approximation generally becomes more accurate.

Applying these methods involves some computation, which is why they are often implemented using computer programs or calculators. This way, they help us evaluate integrals, such as \(\int_{-1}^{0} \frac{e^{\frac{1}{x}}}{x^{3}} d x\), which might be challenging to integrate by hand.

Hyperbolic Tangent Function Integral

The hyperbolic tangent function, often denoted as \(\tanh x\), has an integral that's encountered frequently in calculus. While not immediately obvious, some expressions can be simplified to form a \(\tanh\) or its variants. A classic example is the integral \(\int_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x}} dx\), which resembles the hyperbolic tangent function.

When faced with this integral, using a substitution that transforms the integrand into a function of hyperbolic tangents, we arrive at an elegant and simple result. In the solution provided, the substitution \(u = e^x\) simplifies the integral into a form that's well-known to be \(\frac{\pi}{2}\). This is a neat example of how recognition of integral forms and appropriate substitution can drastically simplify a problem.

This concept is vital as it showcases the power of substitution to handle integrals involving hyperbolic functions, which occur quite often in mathematics and physics scenarios.

Integral of Rational Functions

Integrating rational functions, which are ratios of polynomials, can become intricate depending on the polynomials' degrees and factors. One effective technique for resolving such integrals involves using partial fraction decomposition when the denominator can be factored into linear or irreducible quadratic factors.

Take, for instance, \(\int_{-1}^{1} \frac{dx}{(2-x) \sqrt{1-x^{2}}}\). Upon using the substitution method and partial fractions, this integral transforms into the sum of simpler integrals that can be solved individually. Thanks to this strategy, even complex-looking integrals can often be broken down into more manageable pieces. For example, the above integral simplifies to separable terms that can be integrated to zero.

Note that the application of the theory of residues from complex analysis is another powerful tool when dealing with definite integrals of certain rational functions over infinite intervals or involving periodic integrands.

Substitution Method in Integration

Substitution is a method often used for integrals with compound functions or those in which direct integration is not straightforward. By substituting a part of the integrand with a new variable, the integral might be reduced to a simpler form that is easier to evaluate.

For example, with the integral \(\int_{3}^{5} \frac{x^{2} dx}{\sqrt{(x-3)(5-x)}}\), we can use the substitution \(u = \sqrt{(x-3)(5-x)}\), which simplifies the expression. Calculus students quickly learn that substitution can be both an art and a science, requiring practice and sometimes creativity to spot the switch that will simplify the problem.

Remember, the key to successful substitution is to also transform the differential (in this case \(dx\)) and update the limits of integration when performing definite integrals. This ensures that the entire integral is translated into the new variable, allowing for seamless integration.