Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

a. (a) Compute \([x \ln x-x]^{\prime}\) (b) Compute \(\int_{1}^{e} \ln x d x\) b. Note: \(\sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\) Compute \(\int_{0}^{\pi} \sin ^{2} \theta d \theta\) c. (a) Show that \(\frac{1}{t}+\frac{1}{1-t}=\frac{1}{t(1-t)}\) (b) Compute \(\int \frac{1}{t(1-t)} d t\) d. \(\quad\) (a) Believe: \([\arcsin x]^{\prime}=\frac{1}{\sqrt{1-x^{2}}}\) (b) Compute \(\int_{0}^{1 / 2} \frac{1}{\sqrt{1-x^{2}}} d x\)

Step-by-Step Solution

Answer

(a) \((x \ln x - x)' = \ln x\); (b) \(\int_1^e \ln x \, dx = e - 2\); (b parts): \(\int_0^\pi \sin^2 \theta \, d\theta = \frac{\pi}{2}\); (b parts): \(\int \frac{1}{t(1-t)} \, dt = \ln |t| - \ln |1-t| + C\); (b parts): \(\int_0^{1/2} \frac{1}{\sqrt{1-x^2}} \, dx = \frac{\pi}{6}\).

Step 1: Differentiate the function

To find \([x \ln x-x]'\), use the product rule and constant rule. The derivative of \(x \ln x\) is \((1 \cdot \ln x + x \cdot \frac{1}{x})\), which simplifies to \(\ln x + 1\). The derivative of \(-x\) is \(-1\). Combine terms: \([x \ln x - x]' = \ln x + 1 - 1 = \ln x\).

Step 2: Integrate using substitution

To integrate \(\int_1^e \ln x \, dx\), use integration by parts: let \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = x\). Thus, the integral is \([x \ln x - x]_1^e\), which evaluates to \(e \ln e - e - (1 \cdot \ln 1 - 1) = e - (0 - 1) = e - 1\).

Step 3: Substitute expression into integral

Given \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\), substitute into the integral: \(\int_0^\pi \frac{1 - \cos 2\theta}{2} \, d\theta\). This simplifies to \(\frac{1}{2} \int_0^\pi (1 - \cos 2\theta) \, d\theta\), which splits into two separate integrals.

Step 4: Evaluate trigonometric integrals

Compute \(\frac{1}{2} \int_0^\pi 1 \, d\theta = \frac{1}{2}[\theta]_0^\pi = \frac{\pi}{2}\) and \(\frac{1}{2} \int_0^\pi -\cos 2\theta \, d\theta = 0\) since the integral of \(\cos 2\theta\) over \([0, \pi]\) is zero. Therefore, the result is \(\frac{\pi}{2}\).

Step 5: Simplify sum of fractions

Show \(\frac{1}{t} + \frac{1}{1-t} = \frac{1}{t(1-t)}\) by finding a common denominator \(t(1-t)\), rewriting the fractions: \(\frac{(1-t) + t}{t(1-t)} = \frac{1}{t(1-t)}\).

Step 6: Compute improper integral

For \(\int \frac{1}{t(1-t)} \, dt\), decompose the fraction into partial fractions: \(\frac{1}{t(1-t)} = \frac{1}{t} + \frac{1}{1-t}\). Integrate both: \(\ln |t| - \ln |1-t| + C\).

Step 7: Confirm derivative of arcsin

According to the problem, \([\arcsin x]' = \frac{1}{\sqrt{1-x^2}}\) is given. Hence, no calculation is needed for verification here.

Step 8: Evaluate the definite integral

Evaluate \(\int_0^{1/2} \frac{1}{\sqrt{1-x^2}} \, dx\) using the arcsine antiderivative. This is \([\arcsin x]_0^{1/2}\), which results in \(\arcsin(1/2) - \arcsin(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}\).

Key Concepts

Derivative Calculation

Calculating derivatives is a key concept in calculus. It involves finding the rate of change of a function with respect to its variables. The derivative of a function gives you the slope of the tangent line at any point on the function's graph. For example, in our problem, calculating the derivative of the expression \([x \ln x - x]'\), involves using the **product rule**.

• The product rule states that the derivative of a product of two functions \(u(x)\) and \(v(x)\) is \(u'v + uv'\).
• In this case, if \(u = x\) and \(v = \ln x\), then the derivative of \(x\ln x\) is \(1 \cdot \ln x + x \cdot \frac{1}{x}\).\
• Simplifying gives \(\ln x + 1\). Moreover, the derivative of \(-x\) is \(-1\).
  Combining both results, you have the complete derivative: \([x \ln x - x]' = \ln x + 1 - 1 = \ln x\).

Understanding these rules helps solve various calculus problems that involve rate of change and determining the slopes of curves.

Integration Techniques

Integration is the process of finding the antiderivative or area under the curve of a function. It is the reverse operation of differentiation. There are several techniques used to solve integrals effectively. One significant technique is **integration by parts**, which is essential for handling integrals of products of functions, like in the integral \(\int_1^e \ln x \, dx\). This technique uses the formula:

• \(\int u \, dv = uv - \int v \, du\)

In this problem, we set \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = x\). This results in the integral becoming \([x \ln x - x]_1^e\), simplifying to \(e\ln e - e - (1 \cdot \ln 1 - 1) = e - 1\). Another useful method involves simple algebraic manipulation, such as **substitution** method seen with \(\sin^2 \theta\) using the identity \(\frac{1-\cos 2\theta}{2}\), making integrals of trigonometric functions more manageable.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They are critically important in the simplification of trigonometric expressions and solving integrals involving trigonometric functions. When calculating the integral \(\int_0^\pi \sin^2 \theta \, d\theta\), we can employ the identity:

• \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\)

Using this identity simplifies the integral into a form that's easier to compute. Splitting it, we get:

• \(\frac{1}{2} \int_0^\pi (1 - \cos 2\theta) \, d\theta\)
• This resolves into two integrals: \(\frac{1}{2} \int_0^\pi 1 \, d\theta = \frac{\pi}{2}\)
• and \(\frac{1}{2} \int_0^\pi -\cos 2\theta \, d\theta\), which equals zero.
Thus, the total integral sum results in \(\frac{\pi}{2}\). Knowing these identities and how to utilize them simplifies solving calculus problems involving trigonometric functions.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to integrate. This approach is particularly useful when you encounter integrals involving fractions. For example, in finding \(\int \frac{1}{t(1-t)} \, dt\), we use partial fraction decomposition by expressing the integrand as a sum of simpler fractions:

• Decompose \(\frac{1}{t(1-t)}\) into \(\frac{1}{t} + \frac{1}{1-t}\).

Once decomposed, both fractions are integrated separately. The integral becomes:

• \(\int \frac{1}{t} \, dt - \int \frac{1}{1-t} \, dt = \ln |t| - \ln |1-t| + C\).

This technique simplifies the process of integration, making it easier to handle more complex expressions. By decomposing fractions into their partial components, you can focus on integrating simpler terms.