The shaded section shown is bounded above by y=7x^2.5, on the right by x=1, and below by the x-axis. For the shaded section determine the moment of inertia with respect to the x-axis, Ix ; the moment of inertia with respect to the y-axis, Iy ; the polar moment of inertia, Jo ; and the radius of gyration with respect to the x -axis, Kx. Show work well please. This image displays a graph with an exponential curve, depicting a mathematical relationship between two variables, \(x\) and \(y\). The horizontal axis is labeled "x (m)" and ranges from 0.0 to 1.0. The vertical axis is labeled "y (m)" and ranges from 0 to 7.**Description of the Graph:**- The curve starts at the origin (0,0) and rises steeply as \(x\) approaches 1.0, indicating an increasing rate of change in \(y\).- The shaded area under the curve represents the integral of the function from \(x = 0\) to \(x = 1\). This area can be interpreted as the total change in \(y\) over the interval.- The nature of the curve suggests it could be an exponential function due to the rapid increase in \(y\) values relative to \(x\).This graph visually illustrates the concept of integration and its geometric interpretation as the area under a curve, which is often used in calculus to find accumulated values or total change.

Area, A=dA=γdx∫dA=∫017x2.5dxA=7×x2.52.510A=7×12.5A=2.8 m2 Coordinates of Centroid x x=∫xdAAx=∫00.30.5+0.5x+1-xdy2.8x=∫00.30.5+0.5×y311.5+1-y323dy2.8x=0.8572.8x=0.306 m (iii) Coordinates of centroid y¯ y¯=∫y¯dAA=∫y2(7x2.5)dx1.2y¯=0.3571 m Calculating Moment of inertia in x-direction Ix=∫y2dAIx=∫01y3dxIx=∫01(7x1.5)3 dx Ix=62.363 m4 Since dA=ydxy=3x1.5Therefore dA=3x1.5 Calculating Moment of inertia in y-direction Iy=∫x2dAIy=∫01x2(7x1.5dx)Iy=∫017x3.5dxIy=1.555 m4 (v) Polar moment of inertia Jo Jo=∫r2dAJo=∫(x2+y2)dAJo=∫x2dA+∫y2dAJo=Iy+IxJo=62.363+1.555Jo=63.918 m4 (vi) Radius of gyration Ky Ky=IyA=62.3632.8Ky=4.719 m