Question

$$\text { Evaluate } \lim _{x \rightarrow 2} \frac{\int_{2}^{x} \sqrt{t^{2}+t+3} d t}{x^{2}-4}$$

Short answer

Expression: $$\lim _{x \rightarrow 2} \frac{\int_{2}^{x} \sqrt{t^{2}+t+3} d t}{x^{2}-4}$$ Answer: The limit of the given expression as x approaches 2 is $$\frac{\sqrt{11}}{4}$$

Step-by-step solution

Differentiate the numerator using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) and if a and x are in the domain of F, then $$\frac{d}{d x} \int_{a}^{x} f(t) d t = f(x)$$ In our case, f(t) = $$\sqrt{t^{2}+t+3}$$. So, to differentiate the numerator, we simply apply the Fundamental Theorem of Calculus: $$\frac{d}{d x} \int_{2}^{x} \sqrt{t^{2}+t+3} d t = \sqrt{x^{2}+x+3}$$

Differentiate the denominator

The denominator is a simple algebraic function: $$g(x) = x^{2} - 4$$. The first derivative of g(x) with respect to x is: $$g'(x) = 2x$$

Use L'Hopital's Rule

L'Hopital's Rule states that if the limit is of indeterminate form, then the limit can be computed as follows: $$\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim _{x \rightarrow a} \frac{f'(x)}{g'(x)}$$ In our case, we now have: $$f'(x) = \sqrt{x^{2}+x+3}$$ $$g'(x) = 2x$$ Therefore: $$\lim _{x \rightarrow 2} \frac{\int_{2}^{x} \sqrt{t^{2}+t+3} d t}{x^{2}-4} = \lim _{x \rightarrow 2} \frac{\sqrt{x^{2}+x+3}}{2x}$$

Evaluate the limit

Now we can find the limit as x approaches 2: $$ \lim _{x \rightarrow 2} \frac{\sqrt{x^{2}+x+3}}{2x} = \frac{\sqrt{2^{2}+2+3}}{2(2)} = \frac{\sqrt{11}}{4}$$ So, the limit of the given expression is: $$\lim _{x \rightarrow 2} \frac{\int_{2}^{x} \sqrt{t^{2}+t+3} d t}{x^{2}-4} = \frac{\sqrt{11}}{4}$$

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus plays a crucial role in understanding the relationships between differentiation and integration. It helps us determine the derivative of an integral function, which is key to solving problems like the one above. This theorem consists of two main parts:

• The first part of the theorem tells us that if we have a continuous function and integrate it, we obtain an antiderivative.
• The second part assures us that if we have an antiderivative, we can compute specific values using definite integrals.

In solving the given exercise, we specifically use the part that states: \(\frac{d}{d x} \int_{a}^{x} f(t) d t = f(x)\). This helps us find the derivative of the numerator. So, if \(f(t) = \sqrt{t^2 + t + 3}\), differentiating \( \int_{2}^{x} \sqrt{t^2 + t + 3} \ dt \) gives us \( \sqrt{x^2 + x + 3}\). This understanding is essential to applying L'Hopital's Rule in the next steps.

Derivative

Derivatives measure how a function changes as its input changes. In simple terms, derivatives represent the slope of a function at any given point. They are fundamental in calculus and play an important role in a variety of problem-solving techniques.

For our exercise, we need to differentiate both the numerator and the denominator. The numerator has already been handled by the Fundamental Theorem of Calculus. For the denominator, we have the function \(g(x) = x^2 - 4\). Its derivative, \(g'(x) = 2x\), represents the rate at which \(x^2 - 4\) changes.

• This use of derivatives helps us apply L'Hopital's Rule, which requires the derivatives of both the numerator and the denominator.
• Understanding derivatives allows us to simplify complex problems by finding slopes of tangent lines at specific points.

Grasping the concept of derivatives not only helps with limits but also aids in optimization problems and understanding motion.

Indeterminate Form

An indeterminate form arises when a mathematical expression does not unambiguously define a particular value. Common examples include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and others like \(0 \times \infty\).

In the context of limits, encountering an indeterminate form means we cannot directly evaluate the limit by substitution. This is where L'Hopital's Rule becomes invaluable: it offers a way to resolve indeterminate forms by differentiating the functions in the numerator and denominator.

• In our problem, substituting \(x = 2\) into \(\frac{\int_{2}^{x} \sqrt{t^2+t+3} \, d t}{x^2 - 4}\) produces \(\frac{0}{0}\), an indeterminate form.
• L'Hopital's Rule then allows us to find the limit by differentiating both the numerator and the denominator instead.

Therefore, understanding indeterminate forms helps us recognize when to utilize calculus tools, like L'Hopital's Rule, to find limits that are otherwise hard to compute directly.