Step 1: The given expression involves the limit of an integral. We are tasked with finding the value of this limit.
Step 2: The integral is given as:
\[ \int_4^{f(x)} \frac{2t}{x-1} \, dt \]
Since \( f(x) \) is differentiable and \( f(1) = 4 \), we can apply the Fundamental Theorem of Calculus.
Step 3: First, rewrite the integral as follows:
\[ \int_4^{f(x)} \frac{2t}{x-1} \, dt \]
Notice that as \( x \to 1 \), \( f(x) \to 4 \). Hence, we are interested in the behavior of the integral as \( x \) approaches 1.
Step 4: The integrand has the form \( \frac{2t}{x-1} \), which suggests that the integral evaluates to a result proportional to \( \frac{1}{x-1} \).
Step 5: Differentiating the integral expression with respect to \( x \), we get:
\[ \frac{d}{dx} \left( \int_4^{f(x)} \frac{2t}{x-1} \, dt \right) = \frac{2f(x)}{x-1} \cdot f'(x) \]
Step 6: Substituting \( f(1) = 4 \) and \( f'(1) = 2 \), we evaluate the limit at \( x = 1 \):
\[ \lim_{x \to 1} \frac{2f(x)}{x-1} \cdot f'(x) = 16 \]
Thus, the value of the given expression is \( \boxed{16} \).
Evaluating the Limit $\lim_{x \to 1} \frac{1}{x-1} \int_{4}^{f(x)} 2t \, dt$
We are given a differentiable function $f: \mathbb{R} \to \mathbb{R}$ with $f(1) = 4$ and $f'(1) = 2$. We need to find the value of the limit:
$$ L = \lim_{x \to 1} \frac{1}{x-1} \int_{4}^{f(x)} 2t \, dt $$
Step 1: Evaluate the definite integral
First, evaluate the integral with respect to $t$:
$$ \int_{4}^{f(x)} 2t \, dt = \left[ t^2 \right]_{4}^{f(x)} = (f(x))^2 - 4^2 = (f(x))^2 - 16 $$
Step 2: Substitute the result back into the limit expression
Now, substitute this back into the limit:
$$ L = \lim_{x \to 1} \frac{(f(x))^2 - 16}{x-1} $$
Step 3: Check for indeterminate form
As $x \to 1$, the numerator approaches $(f(1))^2 - 16 = 4^2 - 16 = 16 - 16 = 0$.
As $x \to 1$, the denominator approaches $1 - 1 = 0$.
Since we have the indeterminate form $\frac{0}{0}$, we can use L'Hôpital's Rule.
Step 4: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator with respect to $x$:
Derivative of the numerator: $\frac{d}{dx}((f(x))^2 - 16) = 2f(x)f'(x)$
Derivative of the denominator: $\frac{d}{dx}(x-1) = 1$
Now, apply the limit to the ratio of the derivatives:
$$ L = \lim_{x \to 1} \frac{2f(x)f'(x)}{1} $$
Step 5: Substitute the given values
Substitute the given values $f(1) = 4$ and $f'(1) = 2$:
$$ L = 2f(1)f'(1) = 2(4)(2) = 16 $$
Therefore, the value of the limit is 16.
Final Answer: (A) 16
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
A stationary tank is cylindrical in shape with two hemispherical ends and is horizontal, as shown in the figure. \(R\) is the radius of the cylinder as well as of the hemispherical ends. The tank is half filled with an oil of density \(\rho\) and the rest of the space in the tank is occupied by air. The air pressure, inside the tank as well as outside it, is atmospheric. The acceleration due to gravity (\(g\)) acts vertically downward. The net horizontal force applied by the oil on the right hemispherical end (shown by the bold outline in the figure) is:
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
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- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: