Problem 61 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{0}^{7} \frac{d x}{x+2}$$

Step-by-Step Solution

Answer

The resultant integral value is \[\ln|9|- \ln|2|\].

Step 1: Identify the substitution

The integral function is of the form \(\frac{1}{x+2}\). Let's make a simple substitution and let \(u = x + 2\). Then, calculate \(du\) which is \(dx\).

Step 2: Transform the integrand

Change the integral in terms of \(u\). The expression \(\frac{dx}{x+2}\) will become \(\frac{du}{u}\).

Step 3: Change the limits of integration

Now, the bounds of the integral also need to be changed. Therefore \(u(0) = 0+2 = 2\) and \(u(7) = 7+2 = 9\). So, the integral will become \(\int_{2}^{9} \frac{du}{u}\).

Step 4: Evaluate the integral

Now, the integral \(\int_{2}^{9} \frac{du}{u}\) is in standard form and can be evaluated as the natural log. The integral thus evaluates to: \[\ln|u|\] evaluated from 2 to 9.

Step 5: Evaluate the integral at the limits of integration

Substitute back the values of \(u\) at the lower and upper limits. Difference of the natural log at upper minus lower value. \[\ln|9|- \ln|2|\].

Key Concepts

Indefinite Integral

An indefinite integral is a primary concept in calculus that represents a family of functions which all differ by a constant. When we integrate a function without specified limits of integration, what we're actually finding is an antiderivative.

For instance, suppose we want the indefinite integral of a function, say, \( f(x) = x \). The antiderivative of \( f(x) \) is \( F(x) = \frac{1}{2}x^2+C \) where \( C \) is the constant of integration. This constant is necessary since the derivative of a constant is zero, and any constant could have existed in the original function before we took its derivative. In the context of our exercise, after the \( u \) substitution, the indefinite integral of \( \frac{1}{u} \) would be \( \ln|u|+C \).

Integration by Substitution

Also known as u-substitution, integration by substitution is a technique used to evaluate integrals. Think of it as the reverse chain rule: you identify part of the integrand that, when differentiated, appears elsewhere in the integrand. You then substitute \( u \) for this part and rewrite the integral in terms of \( u \).

This technique simplifies the integrand and makes the integral easier to solve. In our exercise, you chose \( u = x + 2 \) because the derivative of \( x + 2 \) is 1, which matches the \( dx \) part of the integral. Then the problematic \( \frac{1}{x+2} \) becomes the friendlier \( \frac{1}{u} \) after the substitution.

Definite Integral

A definite integral has both an upper and lower limit of integration, and it calculates the net area under a curve between these two points. Unlike an indefinite integral, the definite integral is a number that represents the accumulation of quantities like area, volume, and other physical concepts.

When performing a substitution, the limits of integration must also change to correspond to the new variable. In the exercise, after changing the variable from \( x \) to \( u \), we also adjusted the limits from 0 and 7 to 2 and 9 respectively, based on the \( u \) substitution. When the integral is evaluated with these new limits, it gives a specific value representing the area under the curve of \( \frac{1}{u} \) between \( u = 2 \) and \( u = 9 \) on the \( u \) axis.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828\) is the Euler's number, a mathematical constant that arises in many areas of mathematics. The function \( \ln(x) \) is the inverse of the exponential function \( e^x \).

In integration, the natural logarithm appears when finding the antiderivative of \( \frac{1}{x} \), where \( x \) is greater than zero. This property is what made \( u \) substitution appealing for our exercise: integrating \( \frac{du}{u} \) resulted in \( \ln|u| \) as the antiderivative. When evaluating this at the integration limits \( u = 2 \) and \( u = 9 \) from our substitution, we use this natural logarithm to find the value of \( \ln|9|-\ln|2| \) which gives the solution to our definite integral.