To find the area enclosed by the curves, we need to set up an integral. First, let's express both curves in terms of \( y \) and solve for the intersection points. The first curve is \( y = x^2 - 4x + 4 \) (a parabola) and the second curve is \( y^2 = 16 - 8x \), which is a sideways parabola.
Next, we solve for the intersection points by equating the two curves, and then integrate the difference of the two functions to find the enclosed area.
After solving the integration, the area enclosed by the curves is \( \frac{8}{3} \).
The eccentricity of the curve represented by $ x = 3 (\cos t + \sin t) $, $ y = 4 (\cos t - \sin t) $ is:
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.