41 # Integration## 38. Refer to Figure 14.40Figure 14.40 shows a diagram with the curves:- \( y = x^2 - 5 \)- \( y = 7 - 2x^2 \) The region in question is to the right of the line \( x = 1 \).## 39.Express, in terms of a single integral, the total area of the region to the right of the line \( x = 1 \) that is between the curves \( y = x^2 - 5 \) and \( y = 7 - 2x^2 \). Do not evaluate the integral.## 40.Express, in terms of a single integral, the total area of the region in the first quadrant bounded by the x-axis and the graphs of \( y = x^2 \) and \( y = 3 - x \). Do not evaluate the integral.## 41-56.Find the area of the region bounded by the graphs of the given equations. Use \( x = \) or \( y = \) as needed. Consider horizontal strips for simplicity.- Example: \( y = x^2 + x - 3 \), \( y = x + 3 \), \( y = 0 \)- Note different forms such as: \( x = y^2 - 4y + 4 \) and similar variations for different equations.## 59. Lorenz CurveA Lorenz curve demonstrates income distribution inequality.- Let \( x \) be the cumulative percentage of income recipients - Let \( y \) be the cumulative percentage of income.When equality is achieved, the Lorenz curve is \( y = x \). For example, when \( x = 10\%\), then \( y = 10\%\).### Figure 14.41It shows the Lorenz curve defined by:\[ y = \frac{14}{15}x^2 + \frac{1}{15}x \]- The graph plots cumulative percentages of income against cumulative percentages of recipients.- Note: 30% of people receive only 10.4% of total income, indicating inequality.The degree of deviation from equality is measured by the coefficient of inequality.## 60-68. ExercisesDetermine areas using coordinates, curves, lines, or given inequalities. Calculate the coefficient of inequality for specific Lorenz

The area under the curves is the bounded region formed by the curves.Sketch the curves y1=x2 and y2=2x. Shade the region bounded by the curves.Both the graphs intersect each other at two points. Equate the curves and solve the obtained equation for x. x2=2x⇒x2-2x=0⇒x(x-2)=0⇒x=0 or x-2=0∴x=0 or x=2 The y-values of the points are, y=2·0=0and, y=2·2=4 Therefore, the points are 0, 0 and 2, 4.The limit of the shaded area lies between x=0 and x=2. The area is, ∫02y2 dx-∫02y1 dx=∫022x dx-∫02x2 dx=2·x2202-x3302=x202-x3302=22-0-233+0=4-83=43 sq, units The area is 43 sq. units.

Math

Calculus

In Problems 41–56, find the area of the region bounded by the graphs of the given equations. Be sure to find any needed points of intersection. Consider whether the use of horizontal strips makes the integral simpler than when vertical strips are used. 41. y=x², y= 2r 42. y = x, y = -x+3, y=0