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Chapter 17: Problem 10

Find \(\int_{0}^{0.3} \frac{1}{x^{2}-3 x+2} \mathrm{~d} x\).

Short Answer

Expert verified
Considering the provided step-by-step solution, please find the result of the definite integral: The definite integral of the function ƒ(x) = (1/x^2 - 3x + 2) on the interval [0, 0.3] is approximately -0.3523.

Step by step solution

01

Factor the denominator

First, we need to factor the denominator of the fraction. In this case, we have \(x^2 - 3x + 2\). Factoring it, we get \((x - 1)(x - 2)\).
02

Decompose the fraction into partial fractions

Now we need to decompose the given fraction \(\frac{1}{(x-1)(x-2)}\) into partial fractions. To do this, we can write: $$ \frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} $$ Multiplying both sides by the common denominator \((x - 1)(x - 2)\), we get: $$ 1 = A(x - 2) + B(x - 1) $$ To find the constants A and B, we can use the following system of equations: $$ \begin{cases} 2A - B = 1\\ -A + B = 0 \end{cases} $$ Solve this system for A and B: $$ \begin{cases} A = \frac{1}{3}\\ B = \frac{1}{3} \end{cases} $$ Now, our decomposed fraction becomes: $$ \frac{1}{(x-1)(x-2)} = \frac{1/3}{x-1} + \frac{1/3}{x-2} $$
03

Integrate the partial fractions

We can now integrate each partial fraction independently: $$ \int (\frac{1/3}{x-1} + \frac{1/3}{x-2}) dx = \frac{1}{3} \int \frac{1}{x-1} dx + \frac{1}{3} \int \frac{1}{x-2} dx $$ Both integrals are in the form \(\int \frac{1}{x-a} dx\), which integrates to \(\ln|x-a| + C\). Therefore, we get the following antiderivative: $$ \frac{1}{3}\ln|x-1| + \frac{1}{3}\ln|x-2| + C $$
04

Evaluate the definite integral on the interval [0, 0.3]

Now we need to evaluate the resulting integral over the interval [0, 0.3] using the fundamental theorem of calculus: $$ \int_{0}^{0.3} \frac{1}{x^{2}-3 x+2} dx = (\frac{1}{3}\ln|\text{0.3}-1| + \frac{1}{3}\ln|\text{0.3}-2|) - (\frac{1}{3}\ln|-1| + \frac{1}{3}\ln|-2|) $$ Calculating the definite integral, we obtain the final result: $$ \int_{0}^{0.3} \frac{1}{x^{2}-3 x+2} dx ≈ -0.3523 $$

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