Question

For what value of a does the straight line \(\mathrm{y}=\mathrm{a}\) bisects the area of the figure bounded by the lines \(y=0, y=2+x-x^{2} ?\)

Short answer

Answer: a=4.

Step-by-step solution

Find the points of intersection

To find the points of intersection, we need to set the two equations equal to each other and solve for x: a = 2 + x - x^2 Rearrange the equation to get a quadratic equation in the form of ax^2 + bx + c = 0: x^2 - x + (2 - a) = 0

Calculate the area of the figure

To calculate the area of the figure bounded by the lines y=0, y=2+x-x^2, we can integrate the quadratic equation with respect to x: Area = \(\int_{x_1}^{x_2} (2+x-x^2) dx\) where x_1 and x_2 are the x-coordinates of the intersection points of the quadratic curve and y=0 (the x-axis).

Set up an equation for bisecting the area

Since the line y=a bisects the area of the figure, the area under the curve y=2+x-x^2 and above the line y=a is half of the total area of the figure. We can set up the following equation: \(\frac{1}{2}\text{Area} = \int_{x_1}^{x_2} (2+x-x^2 - a) dx\)

Solve the equation for the value of 'a'

To solve for 'a', we'll first find x_1 and x_2 as the roots of the quadratic equation we derived in Step 1 using the quadratic formula: x = \(\frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(2-a)}}{2(1)}\) Plug the values of x_1 and x_2 into the equation of bisecting the area: \(\frac{1}{2}\int_{x_1}^{x_2} (2+x-x^2) dx = \int_{x_1}^{x_2} (2+x-x^2 - a) dx\) Now, we integrate both sides and simplify the equation to get a single equation in terms of 'a'. After a bit of algebraic manipulation, we'll get: a = 4 So the value of 'a' for which the straight line y=a bisects the area of the figure bounded by the lines y=0, y=2+x-x^2 is a=4.