Question

If $I = \int_{1}^{3} \sqrt{3 + x + x^2} dx$, then $I$ lies in the interval

• A. $(2\sqrt{5}, 2\sqrt{15})$
• B. $(\sqrt{3}, 2\sqrt{5})$
• C. $(\sqrt{23}, \sqrt{33})$
• D. $(2\sqrt{15}, \sqrt{23})$

Hint:

To find the interval for a definite integral without explicitly evaluating it, determine the minimum and maximum values of the integrand on the interval of integration. Then, multiply these bounds by the length of the interval to find the lower and upper bounds for the integral.

Answer 1

The Correct Option is A

Step 1: Analyze the integrand.
Let $f(x) = \sqrt{3 + x + x^2}$. We need to find the bounds of this function on the interval $[1, 3]$.
Step 2: Find the minimum and maximum values of the expression inside the square root.
Consider $g(x) = 3 + x + x^2$. The derivative is $g'(x) = 1 + 2x$. On the interval $[1, 3]$, $g'(x)>0$, so $g(x)$ is an increasing function. The minimum value of $g(x)$ on $[1, 3]$ occurs at $x = 1$: $g(1) = 3 + 1 + 1^2 = 5$ The maximum value of $g(x)$ on $[1, 3]$ occurs at $x = 3$: $g(3) = 3 + 3 + 3^2 = 15$
Step 3: Find the bounds for the integrand.
Since $g(x)$ is between 5 and 15 on $[1, 3]$, the integrand $f(x) = \sqrt{g(x)}$ is between $\sqrt{5}$ and $\sqrt{15}$: $\sqrt{5} \le \sqrt{3 + x + x^2} \le \sqrt{15}$ for $1 \le x \le 3$.
Step 4: Use the property of definite integrals for bounded functions.
If $m \le f(x) \le M$ on the interval $[a, b]$, then $m(b - a) \le \int_{a}^{b} f(x) dx \le M(b - a)$. Here, $m = \sqrt{5}$, $M = \sqrt{15}$, $a = 1$, and $b = 3$, so $b - a = 3 - 1 = 2$. Therefore, $\sqrt{5} \cdot (2) \le \int_{1}^{3} \sqrt{3 + x + x^2} dx \le \sqrt{15} \cdot (2)$ $2\sqrt{5} \le I \le 2\sqrt{15}$ Thus, $I$ lies in the interval $\boxed{(2\sqrt{5}, 2\sqrt{15})}$.