Question

Evaluate the following integrals: (i) \(\int a^{m x} \cdot b^{n x} d x\) (ii) \(\int(2 x+3 x) 2 d x\) (iii) \(\int \frac{\mathrm{e}^{3 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}\) (iv) \(\int \mathrm{e}^{\ln 2+\ln \mathrm{x}} \mathrm{dx}\)

Short answer

(i) \(\int a^{mx} \cdot b^{nx} dx\) (ii) \(\int (2x + 3x)2 dx\) (iii) \(\int \frac{e^{3x}+e^{3x}}{e^x+e^{-x}} dx\) (iv) \(\int e^{\ln 2 + \ln x} dx\) Answer: (i) \(\frac{(a^m \cdot b^n)^x}{\ln{(a^m \cdot b^n)}} + C\) (ii) \(5x^2 + C\) (iii) \(\ln |e^{2x} + 1| + C\) (iv) \(x^2 + C\)

Step-by-step solution

Use the product rule for exponentiation

Recall that \(a^{m x} \cdot b^{n x} = (a^m \cdot b^n)^x\). Rewrite the integral using this rule: $$\int a^{mx} \cdot b^{nx} dx = \int (a^m \cdot b^n)^x dx$$

Integrate using the power rule for exponentials

The power rule for exponential functions states: $$\int c^{kx} dx= \frac{c^{kx}}{\ln{(c)} \cdot k} + C$$ Here, \(c = a^m \cdot b^n\) and \(k = 1\). Therefore, the integral becomes: $$\int (a^m \cdot b^n)^x dx = \frac{(a^m \cdot b^n)^x}{\ln{(a^m \cdot b^n)}} + C$$ Integral (ii): Evaluate \(\int (2x + 3x)2 dx\)

Simplify the integrand

Combine the like terms within the integral: $$\int (2x + 3x)2 dx = \int 5x \cdot 2 dx = \int 10x dx$$

Integrate using the power rule for polynomials

The power rule for polynomials states: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ Here, \(n = 1\). Therefore, the integral becomes: $$\int 10x dx = 10 \cdot \frac{x^{1+1}}{1+1} + C = 5x^2 + C$$ Integral (iii): Evaluate \(\int \frac{e^{3x}+e^{3x}}{e^x+e^{-x}} dx\)

Simplify the integrand

Combine the like terms within the integral: $$\int \frac{e^{3x}+e^{3x}}{e^x+e^{-x}} dx = \int \frac{2e^{3x}}{e^x+e^{-x}} dx$$

Multiply numerator and denominator by \(e^{-x}\)

This simplifies the integral: $$\int \frac{2e^{3x}}{e^x+e^{-x}} dx = \int \frac{2e^{3x}e^{-x}}{(e^x+e^{-x})e^{-x}} dx = \int \frac{2e^{2x}}{e^{2x}+1} dx$$

Let \(u = e^{2x}\)

Here we will use substitution. Let \(u = e^{2x} \implies du = 2e^{2x}dx\): $$\int \frac{2e^{2x}}{e^{2x}+1} dx = \int \frac{du}{u+1}$$

Integrate using the natural logarithm rule

The rule for integrating this form states: $$\int \frac{du}{u+1} = \ln |u+1| + C$$ So, the integral becomes: $$\int \frac{2e^{2x}}{e^{2x}+1} dx = \ln |e^{2x} + 1| + C$$ Integral (iv): Evaluate \(\int e^{\ln 2 + \ln x} dx\)

Simplify the exponent

Combine the logarithms within the exponent: $$\int e^{\ln 2 + \ln x} dx = \int e^{\ln 2x} dx$$

Recognize the inverse of the natural logarithm

Recall that \(e^{\ln a} = a\). Therefore, the integral becomes: $$\int e^{\ln 2x} dx = \int 2x dx$$

Integrate using the power rule for polynomials

As we did in integral (ii), the power rule for polynomials can be applied: $$\int 2x dx = 2 \cdot \frac{x^{1+1}}{1+1} + C = x^2 + C$$

Key concepts

Integration Techniques

Integration techniques are key tools for solving calculus problems involving the calculation of areas, volumes, and other quantities defined by an integral. When approaching an integral, the first step is often to simplify the integrand, if possible. This might involve algebraic manipulation, such as factoring, expanding, or combining like terms.

Next, one of several integration strategies might be applied. Common strategies include substitution, where one replaces a part of the integral with a new variable to simplify the calculation, and integration by parts, which is useful for products of functions. In some cases, one might recognize a function as the derivative of another function and apply the fundamental theorem of calculus directly.

For example, the solution to integral (i) uses the product rule for exponentiation to combine bases, simplifying the integrand into a single exponential form allowing the use of the power rule for exponentials. Integral (iii) is handled by substituting a part of the integrand and using the natural logarithm rule. Each technique chosen is context-specific, aiming to simplify the calculus problem to a manageable form.

Power Rule for Exponentials

The power rule for exponentials is a handy way to integrate functions of the form \(c^{kx}\), where \(c\) is a constant and \(k\) is the coefficient of \(x\) in the exponent. The rule can be stated as \(\int c^{kx} dx= \frac{c^{kx}}{\ln{(c)} \cdot k} + C\), where \(C\) represents the constant of integration.

This formula is derived from the chain rule in differentiation and provides a direct method to integrate exponential functions without much hassle. When applying this rule, it's crucial to remember that \(\ln{(c)}\), the natural logarithm of the constant \(c\), acts as a scaling factor in the denominator. In our example with integral (i), this rule reduced a potentially complex calculation to a straightforward one by converting the integration of an exponential into a division by the natural logarithm of its base.

Power Rule for Polynomials

The power rule for polynomials is an essential tool for integrating polynomial functions. When faced with a polynomial \(x^n\), the power rule tells us that the antiderivative is \(\frac{x^{n+1}}{n+1}\), plus the constant of integration \(C\). This rule simplifies the integration of powers of \(x\) by incrementing the exponent by one and dividing the result by the new exponent.

In the provided exercise, integrals (ii) and (iv) demonstrate how to apply the power rule for polynomials. After simplifying the integrand to a simple polynomial expression, the power rule is used to seamlessly obtain its integral. For example, in integral (ii), the integrand was simplified to \(10x\) before the power rule was applied, yielding \(5x^2 + C\) as the antiderivative.

Exponential Functions

Exponential functions, denoted by \(e^{kx}\) where \(e\) is Euler’s number (approximately 2.71828) and \(k\) is a constant, are fundamental in calculus. They have the unique property that the rate of growth (or decay) of the function is proportional to its current value, making them inherently linked to processes involving growth, decay, or compound interest.

Due to their special exponential nature, the derivative of an exponential function is proportional to the function itself, making them particularly amenable to techniques involving both differentiation and integration. This is seen in integral (iii), where the simplicity of the exponential function allows the use of substitution followed by the application of the natural logarithm rule for integration.

Natural Logarithm

The natural logarithm, denoted as \(\ln{x}\), is the inverse function of the exponential function \(e^x\). This means that \(e^{\ln{x}} = x\) for \(x > 0\). The natural logarithm is fundamental in integration because it provides a way to integrate functions of the form \(\frac{1}{x}\) and variations thereof.

The integral of \(1/x\) is \(\ln| x| + C\), and this relationship extends to more complicated functions where \(x\) is replaced with a function of \(x\). In Integral (iii), we applied substitution to get an integrand of the form \(1/(u+1)\), which integrates to \(\ln|u+1| + C\), showcasing the application of the natural logarithm in integration. The natural logarithm's properties are also used in simplifying exponential functions, as seen in integral (iv) where \(e^{\ln{2x}}\) simplifies to \(2x\) before applying the power rule for polynomials.