Problem 50 - Thomas Calculus

Question

Evaluate each integral in Exercises \(47-52\) by reducing the improper fraction and using a substitution (if necessary) to reduce it to standard form. $$ \int_{-1}^{3} \frac{4 x^{2}-7}{2 x+3} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-4 + 2 \ln 9\).

Step 1: Polynomial Long Division

To simplify the integral, first perform polynomial long division on the integrand \(\frac{4x^2 - 7}{2x + 3}\). Divide \(4x^2 - 7\) by \(2x + 3\). The first term in the quotient is \(2x\), found by dividing the leading term \(4x^2\) by \(2x\). Multiply \(2x\) by \(2x + 3\) to get \(4x^2 + 6x\). Subtract \(4x^2 + 6x\) from \(4x^2 - 7\) to get \(-6x - 7\).

Step 2: Continue Division

Next, divide \(-6x\) by \(2x\) to get \(-3\). Multiply \(-3\) by \(2x + 3\) to obtain \(-6x - 9\). Subtract \(-6x - 9\) from \(-6x - 7\) to get a remainder of \(2\). Thus, \(\frac{4x^2 - 7}{2x + 3} = 2x - 3 + \frac{2}{2x + 3}\).

Step 3: Rewrite the Integral

Substitute the result from polynomial long division into the integral: \(\int_{-1}^{3} \left( 2x - 3 + \frac{2}{2x+3} \right) dx\). This breaks down into three separate integrals: \(\int_{-1}^{3} 2x \, dx\), \(\int_{-1}^{3} -3 \, dx\), and \(\int_{-1}^{3} \frac{2}{2x+3} \, dx\).

Step 4: Evaluate the First Integral

Evaluate \(\int_{-1}^{3} 2x \, dx\). The antiderivative is \(x^2\). Compute \([x^2]_{-1}^{3} = (3)^2 - (-1)^2 = 9 - 1 = 8\).

Step 5: Evaluate the Second Integral

Evaluate \(\int_{-1}^{3} -3 \, dx\). The antiderivative is \(-3x\). Compute \([-3x]_{-1}^{3} = (-3 \times 3) - (-3 \times -1) = -9 + 3 = -12\).

Step 6: Evaluate the Third Integral

Use substitution for \(\int_{-1}^{3} \frac{2}{2x+3} \, dx\). Let \(u = 2x + 3\), then \(du = 2dx\), so \(dx = \frac{1}{2}du\). Change the limits: when \(x = -1\), \(u = 1\); when \(x = 3\), \(u = 9\). The integral becomes \(\int_{1}^{9} \frac{1}{u} \, du\). The antiderivative is \(\ln |u|\). Compute \(2 \cdot [\ln |u|]_{1}^{9} = 2(\ln 9 - \ln 1) = 2 \ln 9\).

Step 7: Combine the Results

Add together the evaluated integrals: \(8 - 12 + 2 \ln 9 = -4 + 2 \ln 9\). Thus, the integral evaluates to \(-4 + 2 \ln 9\).

Key Concepts

Polynomial Long Division

When faced with a complex rational algebraic expression, especially in calculus, simplifying it often involves using a technique called polynomial long division. This method helps break down complicated fractions into simpler terms. In our exercise, the integrand \( \frac{4x^2 - 7}{2x + 3} \) is the target for simplification.

To perform polynomial long division:

• Identify the dividend (\(4x^2-7\)) and the divisor (\(2x+3\)).
• Divide the leading term of the dividend by the leading term of the divisor to find the first term of the quotient, which is \(2x\).
• Multiply \(2x\) by the divisor \(2x + 3\) to get \(4x^2 + 6x\). Subtract this from the original dividend.
• Repeat the process with the new dividend \(-6x - 7\).
• This results in a quotient of \(2x - 3\) and a remainder of \(2\), simplifying our original expression to \(2x - 3 + \frac{2}{2x + 3}\).

Polynomial long division is a powerful tool in integral evaluation as it reduces improper fractions into manageable parts, ready for further calculus techniques like integration.

Substitution Method

Substitution is a versatile method used in calculus to simplify integrals, making them easier to evaluate. This technique essentially changes the variable of integration to simplify the function.

In our problem, after using polynomial long division, we encounter the integral:\( \int_{-1}^{3} \frac{2}{2x+3} \, dx \).

• We choose \( u = 2x + 3 \), which means \( du = 2 \, dx \).
• Rearrange to find \( dx = \frac{1}{2} \, du \).
• Change the limits of integration according to the substitution: when \( x = -1 \), \( u = 1 \); and when \( x = 3 \), \( u = 9 \).
• The integral is transformed into \( \int_{1}^{9} \frac{1}{u} \, du \), a standard logarithmic form.

The substitution method allows us to replace a complex form with a simpler one, by using a new variable \( u \), helping us to easily calculate the integral's value.

Improper Fraction Simplification

Improper fractions, with numerators of higher degree than denominators, can complicate integration. Simplification is key to converting these into friendlier forms.

Starting with \( \frac{4x^2 - 7}{2x + 3} \), we realize it's improper since the degree of the numerator (2) is greater than that of the denominator (1). By using polynomial long division:

• We transform this improper fraction to \( 2x - 3 + \frac{2}{2x+3} \).
• This results in a polynomial part (\(2x - 3\)) and a simple rational part (\(\frac{2}{2x+3}\)).

Simplifying improper fractions is crucial for clean integral evaluation. We essentially break down the fraction to its simplest components, each of which is more manageable to integrate.

Antiderivative

Finding antiderivatives is a core skill in calculus that allows you to evaluate definite integrals. An antiderivative of a function gives us another function, whose derivative returns the original function.

For example, in our integral \( \int_{-1}^{3} 2x \, dx \), the antiderivative of \(2x\) is \(x^2\). Other components include:

• For the integral \( \int_{-1}^{3} -3 \, dx \), the antiderivative is \(-3x\).
• To solve \( \int_{1}^{9} \frac{1}{u} \, du \) (from substitution), the antiderivative is the natural logarithm, \( \ln |u| \).

By finding antiderivatives, you translate a differential equation into a more usable algebraic expression. This step is fundamental in moving from the rate of change to the accumulated value, which is what definite integrals seek to determine.

Definite Integrals

Definite integrals allow us to calculate the exact area under a curve, over a specific interval. They provide a powerful way to find accumulated values, unlike indefinite integrals which include an arbitrary constant.

When evaluating the given exercise, the definite integral is:\( \int_{-1}^{3} \left( 2x - 3 + \frac{2}{2x+3} \right) dx \).

• This splits into three separate integrals for easier computation.
• After calculating the antiderivatives for each part: \( [x^2]_{-1}^{3} = 8 \), \([-3x]_{-1}^{3} = -12 \), and \( 2[\ln |u|]_{1}^{9} = 2 \ln 9 \).
• Combining these results, the overall value of the definite integral becomes \(-4 + 2 \ln 9\).

Definite integrals stabilize results within a specified range, elegantly summing up continuous values into precise numbers.