Question

Let F : [0, 2] → \(\R\) be the function defined by
\(F(x)=\int_{x^2}^{x+2}e^{x[t]}dt,\)
where [t] denotes the greatest integer less than or equal to t. Then the value of the derivative of F at x = 1 equals

• A. e³ + 2e² - e
• B. e³ - e² + 2e
• C. e³ - e + 2e²
• D. e³ + 2e² + e

Answer 1

The Correct Option is C

To find the derivative of the function \(F(x)=\int_{x^2}^{x+2}e^{x[t]}dt\) at \(x = 1\), we will use Leibniz's Rule for the differentiation of integrals. According to Leibniz's rule, if:

\(F(x) = \int_{a(x)}^{b(x)} f(x, t) \, dt\)

then the derivative is:

\(\frac{d}{dx} F(x) = \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) \, dt + f(x, b(x)) \cdot b'(x) - f(x, a(x)) \cdot a'(x)\)

In our case, \(a(x) = x^2\) and \(b(x) = x + 2\), and \(f(x, t) = e^{x[t]}\). Thus, we compute each part:

1. Partial derivative of \(f(x, t)\) with respect to \(x\):
2. Compute the integral of the partial derivative:
3. Evaluate \(f(x, b(x)) \cdot b'(x)\):
4. Evaluate \(f(x, a(x)) \cdot a'(x)\):

Thus, substituting these into Leibniz’s formula:

\(F'(1) = 0 + e^{3} \cdot 1 - 2e\)

Therefore, the value of the derivative at \(x = 1\) is:

\(e^3 - e + 2e^2\)

Hence, the correct answer is: e³ - e + 2e².