What is the area of the region bounded by the parabola x/y 2 2y?

y = x – 2 2 = x – 2 4 = x. of the region is just y = x – 2.

At what point on the curve y E x the slope is 1?

(0,1)
The point at which the slope of the curve y = ex is 1 is (0,1).

What is the area enclosed by the curve 2x 3y 12?

So, area is 8 units.

How do you find the area of a bounded parabola?

Thus, the area between the x – axis and the parabola is \[\dfrac{8\sqrt{2}{{a}^{2}}}{3}\]. Therefore, the area between the straight line and parabola is given as the difference between the area under the line and the area under the parabola.

What is the area between the curves X Y and Y x3 from Y − 1 to Y 1?

Thus the area is ∫21−x2+4x+1−(−x3+7×2−10x+3)dx=∫21×3−8×2+14x−2dx. Figure 9.1. 3. Area between curves.

What is order of differential equation?

The order of a differential equation is defined to be that of the highest order derivative it contains. The degree of a differential equation is defined as the power to which the highest order derivative is raised. The equation (f‴)2 + (f″)4 + f = x is an example of a second-degree, third-order differential equation.

What will be the area formed by the pair of lines 2x 3y 12 & XY 1 0 with Y axis?

⇒ Area = 7.5 Sq.

How do you find the area bounded by a curve and a line?

The area under a curve between two points can be found by doing a definite integral between the two points. To find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b. Areas under the x-axis will come out negative and areas above the x-axis will be positive.

How do you find the area of the region between curves?

Replace x x with 2 2 in the equation. Multiply 2 2 by 2 2. The solution to the system is the complete set of ordered pairs that are valid solutions. The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region.

What is the target area of Y = 4x-x^2?

The area targeted is above the curve, below y =4, and to the right of x = 0. The area Targeted above the curve = Rectangle area – Area below the curve: The answer is 8/3. The graph of y = 4x-x^2 is a parabola, open down, with max at (2,4) and x-intercepts 0 and 4.

How do you find the intersection of two curves?

Solve by substitution to find the intersection between the curves. Tap for more steps… Substitute x 2 x 2 for y y into y = 4 x − x 2 y = 4 x – x 2 then solve for x x. Tap for more steps… Replace y y with x 2 x 2 in the equation.

What is the maximum value of Y in a parabola?

The curve is a parabola with a maximum value of y= 4 at (2,4). The area that is specified is bounded by the lines x =0 (y-axis), y = 4 and the portion of the parabola above the curve to the left, between x = 0 and x = 2 and below the horizontal line y = 4