1. Let the given function be:
\[F(x) = \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} dt.\]
2. Differentiate \(F(x)\) with respect to x using the Leibniz rule:
\[F'(x) = \frac{d}{dx}\left(\int_{0}^{x^2} f(t) dt\right) = f(x^2) \cdot \frac{d}{dx}(x^2) = f(x^2) \cdot 2x.\]
Here,
\[f(t) = \frac{t^2 - 5t + 4}{2 + e^t}.\]
3. For the extremum, set \(F'(x) = 0\):
\[f(x^2) \cdot 2x = 0.\]
This gives two cases:
\(x = 0\) (which is not valid for extremum as it lies on the boundary), \(f(x^2) = 0.\)
4. Solve \(f(x^2) = 0\):
\[\frac{t^2 - 5t + 4}{2 + e^t} = 0 \Rightarrow t^2 - 5t + 4 = 0.\]
5. Factorize \(t^2 - 5t + 4 = 0\):
\[(t - 1)(t - 4) = 0 \Rightarrow t = 1, t = 4.\]
6. Since \(t = x^2\), we get:
\[x^2 = 1 \Rightarrow x = \pm 1\]
\[x^2 = 4 \Rightarrow x = \pm 2.\]
Thus, the points of extremum are \(\pm 1\) and \(\pm 2\).
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,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: