Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{1} \frac{(v+1)(v+2)}{2 v^{3}+9 v^{2}+12 v+36} d v$$

Short answer

Question: Evaluate the definite integral of the given function. $$\int_{0}^{1} \frac{(v+1)(v+2)}{2 v^{3}+9 v^{2}+12 v+36} d v$$ Answer: The definite integral of the given function is $$\int_{0}^{1} \frac{(v+1)(v+2)}{2 v^{3}+9 v^{2}+12 v+36} d v = \ln \left|\frac{31}{33}\right|$$

Step-by-step solution

Choose an appropriate substitution

Let us substitute \(v = u - 1\). This choice is made to simplify the denominator by removing \(9v^2\) term and make it a perfect square. Then, \(dv = du\).

Change the limits of integration

Now, we need to change the limits of integration according to the new variable \(u\). When \(v = 0\), we have \(u = 0 + 1 = 1\). When \(v = 1\), we have \(u = 1 + 1 = 2\). So, the new limits of integration are \(1\) and \(2\).

Rewrite the integrand with the substitution

Substituting \(v = u - 1\) in the given integral, we get: $$\int_{1}^{2} \frac{((u - 1) + 1)( (u - 1) + 2)}{2 (u - 1)^{3} + 9 (u - 1)^{2} + 12 (u - 1) + 36} d u$$ $$= \int_{1}^{2} \frac{u(u+1)}{2 (u^{2} - 2u +1) + 9 (u^{2} - 2u + 1) + 12u - 12 + 36} du$$ $$= \int_{1}^{2} \frac{u(u+1)}{11u^2-15u+37} du$$

Integrate the simplified function

Notice that the numerator is the derivative of the denominator, so we can directly integrate this function using the substitution rule without any further simplification. Let \(w = 11u^2 -15u + 37\). Then, \(dw = (22u -15)du\). Now the integral becomes: $$\int \frac{(u+1)du}{11u^2-15u+37} = \int \frac{dw}{w} =\int \frac{1}{w} dw$$ The integral of \(\frac{1}{w}\) is \(\ln|w|\) plus a constant.

Evaluate the integral within the given limits

We have $$\int_{1}^{2} \frac{u(u+1)}{11u^2-15u+37} du = \Bigg[\ln|11u^2-15u+37|\Bigg]_1^2$$ Now, we plug in the limits: $$\ln|11(2)^2-15(2)+37| - \ln|11(1)^2-15(1)+37|$$ $$= \ln \left|\frac{11(2)^2-15(2)+37}{11(1)^2-15(1)+37}\right|$$ $$= \ln \left|\frac{31}{33}\right|$$

Write the final answer

The definite integral of the given function is $$\int_{0}^{1} \frac{(v+1)(v+2)}{2 v^{3}+9 v^{2}+12 v+36} d v = \ln \left|\frac{31}{33}\right|$$

Key concepts

Change of Variables

The change of variables is a transformative tool used in calculus to simplify the process of finding an integral. When dealing with a complicated function, it often helps to introduce a new variable that makes the expression simpler and easier to integrate. This technique leverages the equivalent transformation of the integral into a function of a new parameter.

In the step-by-step solution, we see a clear example of this technique. Instead of integrating with respect to variable \( v \), we introduce a substitution such as \( v = u - 1 \). This shifts our perspective and sometimes simplifies the denominator or another part of the integrand.

• By substituting \( v = u - 1 \), we aim to create a simpler or more recognizable function.
• The differential element changes accordingly, here simply to \( dv = du \).

Notice how the complex-looking polynomial in the denominator becomes more manageable once it becomes a function of \( u \) rather than \( v \). This substitution paves the way for smoother integration.

Limits of Integration

When performing a change of variables in a definite integral, it's crucial to adjust the limits of integration to reflect the new variable. The limits originally applied to \( v \), but after substitution, they need to correspond to \( u \).

Here's how you update the limits:

• If the initial interval was \([a, b]\), and you have substituted \( v = u - 1 \), you evaluate \( u_1 \) at \( v = a \) and \( u_2 \) at \( v = b \).

In the example given, when \( v = 0 \), the new limit for \( u \) becomes 1 (since \( u = 0 + 1 = 1 \)). Similarly, when \( v = 1 \), \( u \) changes to 2 (as \( u = 1 + 1 = 2 \)). By adjusting the limits from \([0, 1]\) to \([1, 2]\), the integral remains accurate and ready for evaluation in terms of \( u \).

Substitution Method

The substitution method is a fundamental technique in calculus for integrating a wide range of functions. At its core, this method makes the integral more accessible by converting a challenging integrand into a simpler form.

In practice:

• Determine a substitution \( v = g(u) \) that simplifies your integral.
• Express all parts of the integrand in terms of \( u \), including the differential element.

In the given solution, by setting \( v = u - 1 \), the integrand transforms significantly. The substitution simplifies the expression, reducing a complex fraction to a form that matches the derivative of its denominator. This reduction often speeds up integration, as seen when the problem is reduced to \( \frac{1}{w} \), which integrates to \( \ln |w| \).

Using these substitutions, we find that integrals, which at first might seem insurmountable, can be broken down systematically, allowing the problem to be approached strategically.