Question

Let $f(x) = \int \frac{x^2 - 3x + 2}{x^4 + 1} \, dx$, then the function decreases in the interval

• A. $(-\infty,-2)$
• B. $(-2,-1)$
• C. $(1, 2)$
• D. $(2,\infty)$

Hint:

Calculus Tip: Do not attempt to integrate the function first! The Fundamental Theorem of Calculus allows you to bypass complex integration entirely when analyzing the rate of change of an area-accumulation function.

Answer 1

The Correct Option is C

Concept: Calculus - Fundamental Theorem of Calculus and Monotonicity (Decreasing Functions).

Step 1: Find the derivative of the function. The given function $f(x)$ is defined as an indefinite integral. By the First Fundamental Theorem of Calculus, the derivative of an integral simply returns the integrand. Therefore, $f^{\prime}(x) = \frac{x^{2}-3x+2}{x^{4}+1}$.

Step 2: State the condition for a decreasing function. A differentiable function $f(x)$ is strictly decreasing on intervals where its first derivative is strictly less than zero. We set the condition: $f^{\prime}(x)<0$.

Step 3: Set up the inequality. Substitute our derivative into the condition: $\frac{x^{2}-3x+2}{x^{4}+1}<0$.

Step 4: Analyze the denominator to simplify the inequality. Look at the denominator $x^{4}+1$. Because $x^4 \ge 0$ for all real numbers $x$, adding 1 ensures that $x^4 + 1 \ge 1>0$. Since the denominator is always strictly positive, the sign of the entire fraction depends entirely on the numerator. Thus, the inequality simplifies strictly to the numerator: $x^{2}-3x+2<0$.

Step 5: Solve the quadratic inequality for the final interval. Factor the quadratic expression: $(x-1)(x-2)<0$. The critical points are $x=1$ and $x=2$. Testing regions on the number line: - For $x<1$ (e.g., $x=0$), $(0-1)(0-2) = 2>0$. - For $1<x<2$ (e.g., $x=1.5$), $(1.5-1)(1.5-2) = (0.5)(-0.5) = -0.25<0$. - For $x>2$ (e.g., $x=3$), $(3-1)(3-2) = 2>0$. The inequality holds true only between the roots. Therefore, $x \in (1, 2)$. $$ \therefore \text{The function decreases in the interval } (1, 2). $$