Question

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is a positive integer. \(\int x(a x+b)^{n} d x(\text { Hint: } u=a x+b .)\)

Short answer

In conclusion, the integral of \(x(ax + b)^n dx\) can be evaluated as \(\frac{1}{a^2} \left[ \frac{(ax+b)^{n+2}}{n+2} - \frac{b (ax+b)^{n+1}}{n+1} \right] + C\), where C is the constant of integration.

Step-by-step solution

Use the given substitution

According to the hint, we will substitute \(u = ax + b\). Now, differentiate the equation with respect to \(x\) to find \(du\): \(\frac{du}{dx} = a \implies du = a \, dx\)

Rewrite the integral in terms of \(u\)

First, we need to express \(x\) and \(dx\) in terms of \(u\). From the substitution \(u = ax + b\), we can write \(x\) as: \(x=\frac{u-b}{a}\) We already found \(du = a\, dx\), so we can write \(dx\) as: \(dx = \frac{du}{a}\) Now, let's substitute these expressions in the original integral: \(\int x(a x+b)^{n} d x = \int \frac{u-b}{a}(u)^{n} \frac{du}{a}\)

Simplify and evaluate the integral

Now, simplify the integral by factoring out constants and integrating with respect to \(u\): \(\int \frac{u-b}{a}(u)^{n} \frac{du}{a} = \frac{1}{a^2} \int (u^{n+1} - b u^n) du\) Now, integrate each term separately: \(\frac{1}{a^2} \int (u^{n+1} - b u^n) du = \frac{1}{a^2} \left[ \frac{u^{n+2}}{n+2} - \frac{b u^{n+1}}{n+1} \right] + C\)

Revert back to the original variable and simplify

Now, substitute back \(u = ax + b\) in the expression we obtained: \(= \frac{1}{a^2} \left[ \frac{(ax+b)^{n+2}}{n+2} - \frac{b (ax+b)^{n+1}}{n+1} \right] + C\) This is the final expression for the evaluated integral.

Key concepts

substitution method

The substitution method is a powerful tool in calculus integration, especially when dealing with complex integrals. It's like a strategic change of variables that simplifies the integral into a more manageable form. The hint given in our exercise suggested using the substitution \( u = ax + b \). By choosing \( u \) in such a way, we align the integral with a more straightforward algebraic expression.

Here's how it works:

• Identify a part of the integral that can be substituted by a simpler variable \( u \), making the rest easier to integrate.
• Derive \( du \) by differentiating \( u \) with respect to \( x \), which helps in rewriting \( dx \) in terms of \( du \).
• Transform the entire integral from \( x \)-terms to \( u \)-terms so you can integrate with respect to \( u \).

Using substitution is akin to taking a detour on a journey - it might seem longer at first, but it makes the path much smoother. The method fits perfectly when you have products of functions and composite functions in your integral.

definite integrals

Definite integrals offer more than just an area-under-the-curve perspective - they help in determining the exact accumulation of quantities over an interval. When you work with definite integrals, you're evaluating the net change from one point to another along the curve.

When solving a definite integral:

• Substitute the limits of integration from the original variable \( x \) into your transformed variable \( u \). This step is crucial to calculate the actual values.
• Be precise in handling constants and coefficients during substitution, as they affect the integral's outcome significantly.
• After integrating, revert to the original variable to find the solution within the defined limits.

The transformation and integration process might look tedious, but the definite integral delivers a precise, numeric result, encapsulating the total change or accumulation of the function between the given bounds, which is invaluable in real-world applications.

integral calculus

Integral calculus is a branch of mathematics focused on the accumulation of quantities and the areas under and between curves. In essence, it's the reverse process of differentiation, helping us find original functions of given derivatives or rates of change.

The crux of integral calculus revolves around:

• Indefinite Integrals: These represent a family of functions and are expressed with a constant of integration \( C \). They broadly tell us what kind of function would differentiate to the integrand.
• Definite Integrals: Unlike their indefinite counterparts, they deliver a specific numeric value that quantifies an area under a curve or between two curves, considering the exact bounds given.
• Methods such as the substitution method simplify integrals when functions formulation becomes complex, allowing us to look for anti-derivatives more effectively.

Integral calculus is instrumental in several fields, from determining accumulated growth rates in economics to calculating physical quantities like distance, mass, and energy in physics. Its versatility and utility make it a fundamental part of mathematics and science.