Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Compute the following antiderivatives or integrals. If you solve the integral by a substitution, \(u(t)=,\) then identify in writing \(u(t)\) and \(u^{\prime}(t)\) a. \(\int e^{2 t} d t\) b. \(\int \sin (2 t) 2 d t\) c. \(\int \frac{1}{\sqrt{t}} d t\) d. \(\int 3 z^{-1} d z\) e. \(\int(3 \cos z+4 \sin z) d z\) f. \(\int\left(\pi^{2}+e^{2}\right) d z\) g. \(\int \frac{x^{2}}{\sqrt{x}} d x\) h. \(\int \frac{x^{2}+1}{x} d x\) i. \(\int\left(x+\frac{1}{x}\right)^{2} d x\) \(j . \int \frac{1}{t+1} d t\) k. \(\int\left(1+t^{2}\right)^{3} 2 t d t\) l. \(\int \frac{1}{z^{5}} d z\)

Step-by-Step Solution

Answer

a) \(\frac{1}{2}e^{2t} + C\); b) \(-\cos(2t) + C\); c) \(2\sqrt{t} + C\); d) \(3\ln|z| + C\); e) \(3\sin z - 4\cos z + C\); f) \((\pi^2 + e^2) z + C\); g) \(\frac{2}{5}x^{5/2} + C\); h) \(\frac{x^2}{2} + \ln|x| + C\); i) \(\frac{x^3}{3} + 2x - \frac{1}{x} + C\); j) \(\ln|t+1| + C\); k) \(\frac{(1+t^2)^4}{4} + C\); l) \(-\frac{1}{4z^4} + C\).

Step 1: Compute integral (a)

For the integral \( \int e^{2t} \, dt \), we'll use a simple substitution. Let \( u = 2t \). Then \( du = 2 \, dt \) or \( dt = \frac{du}{2} \). The integral becomes \( \int e^{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int e^{u} \, du = \frac{1}{2} e^u + C = \frac{1}{2} e^{2t} + C \).

Step 2: Compute integral (b)

For the integral \( \int \sin(2t) \, 2 \, dt \), observe that \( 2 \cdot \int \sin(2t) \, dt \) is given. We can substitute \( u = 2t \), then \( du = 2 \, dt \) or \( dt = \frac{du}{2} \). This gives \( 2 \cdot \int \sin(u) \cdot \frac{1}{2} \, du = \int \sin(u) \, du = -\cos(u) + C = -\cos(2t) + C \).

Step 3: Compute integral (c)

Let's find the antiderivative of \( \int \frac{1}{\sqrt{t}} \, dt \). Rewrite the integrand as \( \int t^{-1/2} \, dt \). The antiderivative of \( t^n \) is \( \frac{t^{n+1}}{n+1} + C \). Here, \( n = -1/2 \), so the antiderivative is \( \frac{t^{1/2}}{1/2} + C = 2 \sqrt{t} + C \).

Step 4: Compute integral (d)

For \( \int 3z^{-1} \, dz \), this is a basic power rule integration. Integrating gives \( 3 \cdot \ln|z| + C \) because the antiderivative of \( z^{-1} \) is \( \ln|z| \).

Step 5: Compute integral (e)

The integral \( \int (3 \cos z + 4 \sin z) \, dz \) can be separated into two parts: \( 3 \cdot \int \cos z \, dz + 4 \cdot \int \sin z \, dz \). The antiderivative of \( \cos z \) is \( \sin z \) and for \( \sin z \) is \( -\cos z \). Therefore, \( 3 \sin z - 4 \cos z + C \).

Step 6: Compute integral (f)

For \( \int (\pi^2 + e^2) \, dz \), recognize that \( \pi^2 + e^2 \) is a constant. So, its integral is \( (\pi^2 + e^2)z + C \).

Step 7: Compute integral (g)

Rewrite \( \int \frac{x^2}{\sqrt{x}} \, dx \) as \( \int x^{3/2} \, dx \). Apply power rule: \( \frac{x^{5/2}}{5/2} + C = \frac{2}{5} x^{5/2} + C \).

Step 8: Compute integral (h)

Split the integral \( \int \frac{x^2 + 1}{x} \, dx \) into two parts: \( \int x \, dx + \int \frac{1}{x} \, dx \). The antiderivative of \( x \) is \( \frac{x^2}{2} \) and of \( \frac{1}{x} \) is \( \ln|x| \). So, the result is \( \frac{x^2}{2} + \ln|x| + C \).

Step 9: Compute integral (i)

Expand the integrand \( \left(x + \frac{1}{x}\right)^2 \) to \( x^2 + 2 + \frac{1}{x^2} \). Integrate term by term: \( \int x^2 \, dx \) is \( \frac{x^3}{3} \), \( \int 2 \, dx \) is \( 2x \), and \( \int \frac{1}{x^2} \, dx \) is \( -\frac{1}{x} \). Thus, \( \frac{x^3}{3} + 2x - \frac{1}{x} + C \).

Step 10: Compute integral (j)

For \( \int \frac{1}{t+1} \, dt \), use the substitution \( u = t+1 \), then \( du = dt \). The integral becomes \( \int \frac{1}{u} \, du = \ln|u| + C = \ln|t+1| + C \).

Step 11: Compute integral (k)

Use substitution for \( \int (1 + t^2)^3 \, 2t \, dt \). Let \( u = 1 + t^2 \), then \( du = 2t \, dt \). The integral becomes \( \int u^3 \, du = \frac{u^4}{4} + C = \frac{(1 + t^2)^4}{4} + C \).

Step 12: Compute integral (l)

For \( \int \frac{1}{z^5} \, dz \), rewrite as \( \int z^{-5} \, dz \). Use the power rule: \( \frac{z^{-4}}{-4} + C = -\frac{1}{4z^4} + C \).

Key Concepts

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the opposite operation of derivatives. When you find the antiderivative of a function, you are essentially determining a function whose derivative gives you the original function. For example, if the derivative of a function is known, say \( f'(x) = x^2 \), then the antiderivative will be \( f(x) = \frac{x^3}{3} + C \), where \( C \) is a constant. This constant is crucial as antiderivatives are not unique; any constant added to an antiderivative still satisfies the equation.

To find antiderivatives:

• Identify the function you are integrating.
• Use known integration rules, like the power rule.
• Add the constant of integration, \( C \), to your result.

Understanding antiderivatives forms the foundation for solving many calculus problems, especially when dealing with definite integrals and real-world applications where restoring the original function from its rate of change is essential.

Substitution method

The substitution method is a clever technique used to simplify complicated integrals. It is especially useful when an integral contains a function and its derivative. The process involves changing the variable of integration to make the integral easier to solve. This change of variable is similar to the use of the chain rule in differentiation.

Here is a step-by-step guide for substitution:

• Select a substitution \( u = g(t) \), where \( g(t) \) is a part of the integrand that makes the integration easier.
• Calculate \( du = g'(t) \, dt \) and solve for \( dt \).
• Replace all instances of \( t \) and \( dt \) in the integral with \( u \) and \( du \).
• Integrate with respect to \( u \).
• Finally, substitute back \( u = g(t) \) to express your answer in terms of the original variable.

For example, in the integral \( \int e^{2t} \, dt \), use \( u = 2t \). This simplifies the integral to one involving \( e^u \), a much more manageable expression.

Power rule

The power rule is a straightforward approach for finding antiderivatives of polynomial expressions. When the expression is in the form \( x^n \), the power rule makes it simple to find its antiderivative. This rule provides a clear formula: if \( n eq -1 \), the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

Here's how you can use the power rule:

• Identify the power \( n \) in \( x^n \).
• Apply the formula \( \frac{x^{n+1}}{n+1} \).
• Add the constant of integration \( C \).

The power rule is applicable to nearly all polynomial functions and is very helpful for integrals where the base is a variable raised to a power. It simplifies the process significantly, allowing you to quickly find antiderivatives, as shown in examples like \( \int x^{3/2} \, dx \). Integrating gives \( \frac{2}{5} x^{5/2} + C \).

Trigonometric integration

Trigonometric integration involves finding the antiderivatives of trigonometric functions. These often require specific techniques and substitutions. Common functions include \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \), and each has standard integration results.

Here are some quick tips:

• The integral of \( \sin(x) \) is \( -\cos(x) + C \).
• The integral of \( \cos(x) \) is \( \sin(x) + C \).
• Make use of substitutions for functions like \( \sin(2x) \), converting to a simpler form.

For example, in solving \( \int \sin(2t) \, 2 \, dt \), notice the presence of \( 2 \) as the derivative of \( 2t \). Using \( u = 2t \) simplifies the integral considerably. Understanding these integrations is essential in calculus as trigonometric functions are prevalent in problems related to periodic phenomena, waves, and circular motion.