
Step 1: Rewrite the function \( y = x|x| \)
The function \( y = x|x| \) can be expressed as: \[ y = \begin{cases} -x^2, & x < 0 \\ x^2, & x \geq 0 \end{cases} \]
Step 2: Graph the function
The graph of \( y = x|x| \) is a parabola, concave downwards for \( x < 0 \) and concave upwards for \( x \geq 0 \). (Refer to the attached graph.)
Step 3: Area computation using integration
The area of the shaded region between \( x = -2 \) and \( x = 2 \) is given by: \[ \text{Area} = \int_{-2}^{2} |y| \, dx = 2 \int_{0}^{2} x^2 \, dx. \]
Step 4: Evaluate the integral
\[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}. \] Thus, the total area is: \[ \text{Area} = 2 \times \frac{8}{3} = \frac{16}{3}. \] Step 5: Final result
The area bounded by the curve \( y = x|x| \), the X-axis, and the ordinates \( x = -2 \) and \( x = 2 \) is: \[ \boxed{\frac{16}{3}}. \]
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).