Partial Fractions Average B23181

1. **Problem Statement:** We are given the function $$f(x) = \frac{3x+5}{x^2+3x+2}$$ defined on the interval $$0 \leq x \leq 1$$. We need to: a) Express $$f(x)$$ in partial fractions. b) Find the average value of $$f(x)$$ on $$[0,1]$$. --- 2. **Part (a): Partial Fraction Decomposition** - The denominator is a quadratic: $$x^2 + 3x + 2$$. - Factor the denominator: $$x^2 + 3x + 2 = (x+1)(x+2)$$ - We express $$f(x)$$ as: $$\frac{3x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$ - Multiply both sides by $$ (x+1)(x+2) $$: $$3x + 5 = A(x+2) + B(x+1)$$ - Expand the right side: $$3x + 5 = A x + 2A + B x + B$$ $$3x + 5 = (A + B) x + (2A + B)$$ - Equate coefficients of like terms: For $$x$$: $$3 = A + B$$ For constants: $$5 = 2A + B$$ - Solve the system: From $$3 = A + B$$, we get $$B = 3 - A$$. Substitute into $$5 = 2A + B$$: $$5 = 2A + (3 - A) = A + 3$$ $$A = 5 - 3 = 2$$ Then, $$B = 3 - 2 = 1$$ - So the partial fraction decomposition is: $$f(x) = \frac{2}{x+1} + \frac{1}{x+2}$$ --- 3. **Part (b): Average Value of $$f(x)$$ on $$[0,1]$$** - The average value of a function $$f$$ on $$[a,b]$$ is given by: $$\text{Average} = \frac{1}{b - a} \int_a^b f(x) \, dx$$ - Here, $$a=0$$ and $$b=1$$, so: $$\text{Average} = \int_0^1 f(x) \, dx$$ - Using the partial fractions: $$\int_0^1 f(x) \, dx = \int_0^1 \left( \frac{2}{x+1} + \frac{1}{x+2} \right) dx = \int_0^1 \frac{2}{x+1} dx + \int_0^1 \frac{1}{x+2} dx$$ - Integrate each term: $$\int \frac{2}{x+1} dx = 2 \ln|x+1| + C$$ $$\int \frac{1}{x+2} dx = \ln|x+2| + C$$ - Evaluate definite integrals: $$\int_0^1 \frac{2}{x+1} dx = 2[\ln(1+1) - \ln(0+1)] = 2(\ln 2 - \ln 1) = 2 \ln 2$$ $$\int_0^1 \frac{1}{x+2} dx = \ln(1+2) - \ln(0+2) = \ln 3 - \ln 2$$ - Sum the results: $$\int_0^1 f(x) \, dx = 2 \ln 2 + (\ln 3 - \ln 2) = \ln 2 + \ln 3 = \ln(2 \times 3) = \ln 6$$ - Since $$b - a = 1 - 0 = 1$$, the average value is: $$\boxed{\ln 6}$$ --- **Final answers:** - a) $$f(x) = \frac{2}{x+1} + \frac{1}{x+2}$$ - b) Average value on $$[0,1]$$ is $$\ln 6$$.