Question

$$ \text { Evaluate the definite integral. } $$ $$ \int_{0}^{1} \frac{3}{2 x^{2}+5 x+2} d x $$

Short answer

From the steps and evaluations, the final answer to this integral is \( - 2*ln(3) + ln(2)\)

Step-by-step solution

Simplify the Function

First, make the denominator of function in simpler terms by applying partial fraction decomposition. Factorize the denominator \(2x^{2} + 5x + 2\). This results in \(2x^{2} + 5x + 2 = (2x+1)(x+2)\). So original function simplifies to \(\frac{3}{(2x+1)(x+2)}\). With partial fraction decomposition, this further simplifies to \(\frac{A}{2x+1} + \frac{B}{x+2}\), where A and B are constants to be determined.

Solve for constants A and B

Clear the fractions by multiplying through by the common denominator to have \(3 = A(x+2) + B(2x+1)\). Setting \(x= -2\) will make A disappear, leaving only B. We can calculate B as follows: \( B = \frac{3}{3}=1\). Similarly, setting \(x= -\frac{1}{2}\) will make B disappear, leaving only A which can be calculate as: \(A= \frac{3}{-1}= -3\). Thus the original function simplifies to \(\frac{-3}{2x+1} + \frac{1}{x+2}\). We can now integrate this term by term.

Evaluate the Integral

Evaluate the integral term by term from 0 to 1.\n \[ \int_0^1 \frac{-3}{2x+1} dx + \int_0^1 \frac{1}{x+2} dx \]\n Using the integral rule \(\int \frac{1}{u} du = ln|u|\), our integral becomes \(-3ln|2x+1| + ln|x+2|\). Now we can use Fundamental Theorem of Calculus and apply the limits from 0 to 1.

Apply the Limits and Simplify

Now apply the upper and lower limits to the expression \(-3ln|2x+1|+ln|x+2|\), this results in the following expression, \[-3ln|3| - 0 + ln|3| - ln|2| = - 2*ln(3) + ln(2)\]

Key concepts

Partial Fraction Decomposition

In calculus, when dealing with integrals involving rational functions, partial fraction decomposition is a powerful technique. It allows us to simplify these functions into a sum of simpler fractions, which are easier to integrate.

For the integral \(\int_{0}^{1} \frac{3}{2 x^{2}+5 x+2} d x\), the first step involved factoring the denominator. By decomposing \(2x^{2} + 5x + 2\), we factor it into \((2x+1)(x+2)\). This factorization reveals that the function can be expressed as a sum of simpler fractions using partial fraction decomposition:

• \(\frac{3}{(2x+1)(x+2)} = \frac{A}{2x+1} + \frac{B}{x+2}\)

Determining the constants \(A\) and \(B\) requires us to clear the fractions by multiplying through by the common denominator, leading to the equation \(3 = A(x+2) + B(2x+1)\). By strategically selecting values for \(x\), we can isolate and solve for \(A\) and \(B\). Setting \(x = -2\) allows us to find \(B\), and setting \(x = -\frac{1}{2}\) solves for \(A\). Thus, the original function is rewritten as a sum of its decomposed parts. This preparation readies the function for integration.

Integral Evaluation

After decomposing the function into its partial fractions, integral evaluation begins. The integral becomes the sum of the integrals of these simpler fractions:

\[ \int_0^1 \frac{-3}{2x+1} dx + \int_0^1 \frac{1}{x+2} dx \]

Each of these integrals follows the rule \(\int \frac{1}{u} du = \ln|u|\), which is a standard result for integrating functions of this form. By applying this rule to each term, we see:

• For \(\int_0^1 \frac{-3}{2x+1} dx\): This results in \(-3 \ln|2x+1|\).
• For \(\int_0^1 \frac{1}{x+2} dx\): This results in \(\ln|x+2|\).

Now that we have the antiderivatives, the next step is to evaluate the definite integral by applying the upper and lower limits.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, providing a way to evaluate definite integrals. It states that if \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then:

\[ \int_a^b f(x) \, dx = F(b) - F(a) \]

In our problem, now that we have the antiderivative as \(-3ln|2x+1| + ln|x+2|\), we apply this theorem from 0 to 1:

• Calculate \(-3\ln|2 \cdot 1 + 1| + \ln|1 + 2|\).
• Subtract the result of \(-3\ln|2 \cdot 0 + 1| + \ln|0 + 2|\).

After performing these calculations, the result simplifies neatly to \(-2 \cdot \ln(3) + \ln(2)\). This step-by-step approach illustrates how these core calculus concepts seamlessly work together to provide an elegant solution to the integral problem.