Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the following indefinite integrals. Check your answers. (a) \(\int 3 \sin (5 t) d t\) (b) \(\int \pi \cos (\pi t) d t\) (c) \(\int \sqrt{3 x+5} d x\) (d) \(\int \frac{\pi}{e^{x}} d x\) (e) \(\int e^{-3 t} d t\) (f) \(\int \sqrt{e^{t}} d t\) (g) \(\int \frac{6}{\sqrt{t^{3}}} d t\) (h) \(\int \frac{1}{3 t+8} d t\)

Step-by-Step Solution

Answer

(a) \(-\frac{3}{5} \cos(5t) + C\) (b) \(\sin(\pi t) + C\) (c) \(\frac{4}{9} (3x + 5)^{\frac{3}{2}} + C \) (d) \(-\pi e^{-x}+C\) (e) \(-\frac{1}{3}e^{-3t} + C \) (f) \(2e^{t⁄2} + C\) (g) \(\frac{-4}{\sqrt{t}} + C\) (h) \( \frac{1}{3} \ln|3t+8| + C\)

Step 1: Part (a)

The integral \(\int 3 \sin (5 t) d t\) can be solved by using substitution where you let \(u = 5t\). The integral then simplifies into \(\frac{3}{5} \int \sin u du\), which results in \(-\frac{3}{5} \cos u + C\). Re-substitute \(5t\) for \(u\) to finalise the solution as \(-\frac{3}{5} \cos(5t) + C\).

Step 2: Part (b)

The integral \(\int \pi \cos (\pi t) d t\) can be solved directly. The antiderivative of cos(\(\pi t\)) is \(\frac{1}{\pi}\) sin(\(\pi t\)) multiplied by the constant \(\pi\), the result is sin(\(\pi t\)) + C.

Step 3: Part (c)

For the integral \(\int \sqrt{3 x+5} d x\), you have to use substitution. Here, you let \(u = 3x+5\), then du = 3 dx. The integral simplifies into \(\frac{2}{3} \int \sqrt{u}du\) which results in \(\frac{2}{3} * \frac{2}{3}(u)^{\frac{3}{2}} + C\). Re-substitute \(3x+5\) for \(u\) to finalise the solution as \(\frac{4}{9} (3x + 5)^{\frac{3}{2}} + C \).

Step 4: Part (d)

For the integral \(\int \frac{\pi}{e^{x}} d x\), the antiderivative of \( \frac{1}{e^{x}} \) is \( -e^{-x} \). So the solution is \(-\pi e^{-x}+C\).

Step 5: Part (e)

The integral \(\int e^{-3 t} d t\) can be solved by using substitution where \(u = -3t\). The integral simplifies into \( -\frac{1}{3} \int e^{u} du\), which leads to \(-\frac{1}{3}e^{u} + C\). Re-substitute \(-3t\) for \(u\) to finalise the solution as \(-\frac{1}{3}e^{-3t} + C \).

Step 6: Part (f)

The integral \(\int \sqrt{e^{t}} d t\) can be re-written as \(\int e^{\frac{t}{2}} d t\). Following standard result for integrals in the form of \(\int e^{kt} dt\), which is \(\frac{1}{k}e^{kt} + C\), the result is \(2e^{t⁄2} + C\).

Step 7: Part (g)

For \(\int \frac{6}{\sqrt{t^{3}}} d t\), rewrite it as \(\int 6t^{-\frac{3}{2}} dt\). Following the power rule in reverse, the solution becomes \(\frac{-4}{\sqrt{t}} + C\).

Step 8: Part (h)

For \(\int \frac{1}{3 t+8} d t\), using substitution where \(u = 3t + 8\), then \(du = 3dt\). The integral simplifies into \( \frac{1}{3} \int \frac{1}{u} du\), which is \( \frac{1}{3} \ln|u| + C\). Replace \(u\) by \(3t+8\), the result is \( \frac{1}{3} \ln|3t+8| + C\).

Key Concepts

Integration Techniques

Grasping the concept of integration techniques is key to solving indefinite integrals, which essentially seek to find the original function, or antiderivative, from its rate of change. It's similar to piecing together the journey by only seeing the steps taken. There are various methods to approach integration, each serving as a strategic tool to simplify complex functions into a more manageable form.

One common method is the Power Rule, where for any function of the form \(x^n\), the integral is \(\frac{x^{n+1}}{n+1}\), provided that \(n \eq -1\). Another technique is Integration by Parts, useful when integrating the product of two functions. It relies on the formula \(\int u dv = uv - \int v du\).

For functions that involve a composition of functions or require factoring, Trigonometric Integration and Partial Fractions come in handy. The former involves using trigonometric identities, while the latter deconstructs rational functions into simpler fractions. Each technique demands a strong foundation in algebraic manipulation and an understanding of function behavior. The goal is to transform the function into a basic form where its antiderivative can be recognized and applied.

U-Substitution

U-substitution is a method resembling the reverse of the chain rule in differentiation, and it is primarily used to handle integrals of composite functions. The idea is to simplify an integral by substituting part of the integrand with a new variable, commonly denoted as \(u\). This process can make a complex-looking integral more approachable.

For effective u-substitution, the following steps are usually taken: First, identify a segment of the integrand to substitute that, once differentiated, appears elsewhere in the integral. Next, express this segment as \(u\), and find the differential \(du\). Now, rewrite the integral in terms of \(u\), and perform the integration on this transformed equation. Finally, reverse the substitution to return the result in terms of the original variable.

Consider the substitution \(u = 5t\) as seen in the exercise's part (a). We notice that \(du/dt = 5\), and it allows the integral to be expressed in a simpler form. After integrating with respect to \(u\), we then replace \(u\) back with \(5t\) to complete the solution. This technique often unveils a recognizable antiderivative that would have been difficult to spot in the integral's original form.

Antiderivatives

Antiderivatives are at the heart of indefinite integrals. They represent a broad family of functions that, when differentiated, yield the original integrand. To put it simply, finding an antiderivative is akin to uncovering the origin story of a rate of change—looking for the formula of the journey itself from the velocity.

The antiderivative of a given function is not unique; it includes a constant of integration \(C\), reflective of the function's infinite possibilities offset by a constant value. For example, the antiderivative of \(2x\) is \(x^2 + C\), where \(C\) can be any real number. This is a consequence of the differential of a constant being zero—adding a constant doesn't alter the rate of change obtained by the original function.

Determining antiderivatives is essentially working backwards from differentiation, but some functions have known antiderivatives that can be directly applied. For instance, the function \(e^x\) is unique as its antiderivative is itself, \(e^x + C\). Recognizing and memorizing the basic antiderivatives, such as polynomials, trigonometric functions, and exponentials, can streamline the integration process notably.