Problem 1 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{0}^{2} x\left(x^{2}-1\right)^{3} d x$$

Step-by-Step Solution

Answer

The evaluated value of the definite integral \(\int_{0}^{2} x(x^{2}-1)^{3} dx\) is \(-\frac{50265}{4}\).

Step 1: Identify u and dv

In this problem, let's identify our functions u and dv as: \(u = x\) and \( dv = (x^2 - 1)^3 dx\)

Step 2: Compute du and v

Now, we find the derivatives and integrals of u and dv respectively: \(du = \frac{d}{dx}(x)dx = dx\) \(v = \int (x^2 - 1)^3 dx\) Now, to find the integral of (x^2 - 1)^3, let's use substitution method: Let t = x^2 - 1, then differentiating t with respect to x, we have: \(\frac{dt}{dx} = 2x\) \(dx=\frac{dt}{2x}\) Now, write the integral in terms of t and perform the integral: \(v = \int t^3 \frac{dt}{2x}\) Now, let's proceed to the next step.

Step 3: Compute the integral using integration by parts

Now we plug u, du, v and dv into the integration by parts formula: \(\int_{0}^{2} x(x^{2}-1)^{3} dx = \int u dv = uv - \int v du\) \(= \int_{0}^{2} x(x^{2}-1)^{3} dx = xv -\int_{0}^{2} v du\) \(=x \int(x^2 - 1)^3 dx - \int_{0}^{2}(\int t^3 \frac{dt}{2x}) dx\) Now, solving the integral in the second part with substitution: \(z = x^2 -1\), then \(dz = 2x dx\) Substitute and solve the integral: \(=-\frac{1}{2} \int_{0}^{2} (\int t^3 dt) dz\) Now, integrating with respect to t and z: \(=-\frac{1}{4}(t^4)\bigg|_{0}^{2}z\) \(=-\frac{1}{4}(z^2 - 1)^4 z\) \(=-\frac{1}{4}((x^2 - 1)^2 - 1)^4 (x^2 -1)\) So, our integral becomes: \(\int_{0}^{2} x(x^{2}-1)^{3} dx = x\int(x^2 - 1)^3 dx - \frac{1}{4}((x^2 - 1)^2 - 1)^4 (x^2 -1)\)

Step 4: Evaluate the definite integral

Now, evaluate the definite integral from 0 to 2: \(\int_{0}^{2} x(x^{2}-1)^{3} dx = \bigg[x\int(x^2 - 1)^3 dx - \frac{1}{4}((x^2 - 1)^2 - 1)^4 (x^2 -1) \bigg] \bigg|_{0}^{2}\) We can now plug the bounds, 0 and 2, into the expression for the integral and subtract the results: \(=(2\int(3)^3 - \frac{1}{4}(15)^4(3) ) - (0\int(-1)^3 dx - \frac{1}{4}(0)^4(-1))\) Now simplify and calculate the expression: \((54 - \frac{50625}{4}) + 0\) \(\int_{0}^{2} x(x^{2}-1)^{3} dx= -\frac{50265}{4}\) So, the evaluated value of the definite integral is -\(\frac{50265}{4}\).

Key Concepts

Integration by Parts

Integration by parts is an essential technique used to simplify complex integrals. It is particularly effective when dealing with products of functions. The theorem is derived from the product rule for differentiation and can be expressed as follows: \\[ \int u \, dv = uv - \int v \, du \] \In layman's terms, you need to select parts of the integral expression that can simplify potential solutions. For example, if you have an integral of the form \(\int x \cdot (some \, function) \, dx\), it might be beneficial to let \(u = x\) since the derivative \(du\) is a simple \(dx\). The integration by parts is particularly useful when differentiating one part simplifies it while integrating the other provides a function we know how to handle. Notably, choosing \(u = x\) is beneficial when differentiating to \[ \frac{d}{dx}(x) = 1\], making further steps easier and straightforward.

Substitution Method

The substitution method, or 'u-substitution,' is a powerful tool for solving integrals by simplifying the integration process. It's especially helpful for integrals formed by functions where one is the derivative of the other one. The important part here is the change of variables, substituting a part of the integral with a simpler variable, say \(t\) or \(z\). \Suppose you have an integral involving a composite function, such as \(\int (x^2 - 1)^3 \, dx\). By substituting \( t = x^2 - 1 \), we simplify the integral to \( \int t^3 \, dt \). This step relies on transforming the original variable \(x\) to \(t\), which usually involves recognizing a chain rule derivative inside the integral. By making use of \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \), we effectively reshape the integral boundaries and functions, resulting in a more straightforward form.

Integration Techniques

In calculus, various integration techniques help solve different types of integrals. Having a toolbox for integration allows tackling problems that might seem complicated at first glance: \

• Basic Integration Rules: The foundation of solving integrals, knowing formulas like constant multiple rule \(\int a \cdot f(x)\, dx = a \cdot \int f(x)\, dx\).
• Integration by Parts: Useful when dealing with products of algebraic and transcendental functions, often converting a complex integral into simpler parts.
• Substitution Method: Handy for integrals involving a function and its derivative or simplifying complex functions.
• Partial Fractions: Breaks down rational functions into simpler fractions that are easier to integrate.

\Understanding when and how to use these techniques is crucial for solving definite integrals efficiently. A strategic approach often involves a combination of these methods, first simplifying before narrowing down to specific techniques like substitution or integration by parts, as seen in our original exercise.