Question

The value of \( \lim_{h \to 0} \left(\frac{1}{h} \int_{4}^{4+h} e^{t^2} dt \right) \) is

• A. \(e^{16}\)
• B. \(e^4\)
• C. \(e^{64}\)
• D. \(e^8\)

Hint:

Recognize the structure \(\lim_{h \to 0} \frac{1}{h} \int_{a}^{a+h} f(t) dt\) as a classic application of the Fundamental Theorem of Calculus. The answer is always just \(f(a)\). In this problem, \(f(t) = e^{t^2}\) and \(a=4\), so the answer is \(f(4) = e^{4^2} = e^{16}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This limit is in the form of the definition of a derivative. Specifically, it relates to the Fundamental Theorem of Calculus (Part 1), which connects differentiation and integration. The theorem states that if \( F(x) = \int_{a}^{x} f(t) dt \), then \( F'(x) = f(x) \).

Step 2: Key Formula or Approach:
The expression can be interpreted using the definition of the derivative: \[ F'(a) = \lim_{h \to 0} \frac{F(a+h) - F(a)}{h} \] Alternatively, we can use L'Hôpital's rule on the \(\frac{0}{0}\) form.

Step 3: Detailed Explanation:
Method 1: Using the Fundamental Theorem of Calculus
Let \(f(t) = e^{t^2}\) and define a new function \(F(x) = \int_{4}^{x} f(t) dt = \int_{4}^{x} e^{t^2} dt\).
According to the Fundamental Theorem of Calculus, the derivative of \(F(x)\) is \(F'(x) = f(x) = e^{x^2}\).
The given limit can be rewritten as: \[ \lim_{h \to 0} \frac{1}{h} \left[ \int_{4}^{4+h} e^{t^2} dt \right] = \lim_{h \to 0} \frac{\int_{4}^{4+h} e^{t^2} dt}{h} \] Using our definition of \(F(x)\), the integral is \(F(4+h) - F(4)\): \[ \lim_{h \to 0} \frac{F(4+h) - F(4)}{h} \] This is exactly the definition of the derivative of \(F(x)\) at the point \(x=4\), i.e., \(F'(4)\).
Since we know \(F'(x) = e^{x^2}\), we can evaluate it at \(x=4\): \[ F'(4) = e^{4^2} = e^{16} \] Method 2: Using L'Hôpital's Rule
As \(h \to 0\), the integral \(\int_{4}^{4+h} e^{t^2} dt \to \int_{4}^{4} e^{t^2} dt = 0\). The denominator is also \(h \to 0\). So we have the indeterminate form \(\frac{0}{0}\).
We can apply L'Hôpital's rule by differentiating the numerator and the denominator with respect to \(h\).
Denominator: \(\frac{d}{dh}(h) = 1\).
Numerator: Using Leibniz integral rule (a part of the Fundamental Theorem of Calculus): \[ \frac{d}{dh} \left( \int_{4}^{4+h} e^{t^2} dt \right) = e^{(4+h)^2} . \frac{d}{dh}(4+h) = e^{(4+h)^2} . 1 \] So the limit becomes: \[ \lim_{h \to 0} \frac{e^{(4+h)^2}}{1} = e^{(4+0)^2} = e^{4^2} = e^{16} \]
Step 4: Final Answer:
Both methods show that the value of the limit is \(e^{16}\).