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Chapter 3: Problem 7

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

Expert verified
Answer: a=4.

Step by step solution

01

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
02

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).
03

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: \(\frac{1}{2}\text{Area} = \int_{x_1}^{x_2} (2+x-x^2 - a) dx\)
04

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.

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Most popular questions from this chapter

Construct the graph of the following curves : (i) \(y^{2}=8 x^{2}-x^{4}\) (ii) \(y^{2}=(x-1)(x+1)^{-1}\).

(a) If \(f(y)=-y^{2}+y+2\), sketch the region bounded by the curve \(x=f(y)\), the \(y\)-axis, and the lines \(\mathrm{y}=0\) and \(\mathrm{y}=1\). Find its area. (b) Find the area bounded by the curve \(x=-y^{2}+\) \(\mathrm{y}+2\) and the \(\mathrm{y}\)-axis. (c) The equation \(\mathrm{x}+\mathrm{y}^{2}=4\) can be solved for \(\mathrm{x}\) as a function of \(\mathrm{y}\), or for \(\mathrm{y}\) as plus or minus a function of \(x\). Sketch the region in the first quadrant bounded by the curve \(x+y^{2}=4\) and find its area first by integrating a function of \(\mathrm{y}\) and then by integrating a function of \(\mathrm{x}\).

Express with the aid of an integral the area of a figure bounded by : (i) The coordinate axes, the straight line \(\mathrm{x}=3\) and the parabola \(\mathrm{y}=\mathrm{x}^{2}+1\). (ii) The \(x\)-axis, the straight lines \(x=a, x=b\) and the curve \(y=e^{x}+2(b>a)\).

At what values of the parameter \(a>0\) is the area of the figure bounded by the curves \(x=a, y=2^{x}\), \(y=4^{x}\) larger or equal to the area bounded by the curves \(\mathrm{y}=2^{\mathrm{x}}, \mathrm{y}=0, \mathrm{x}=0, \mathrm{x}=\mathrm{a} ?\)

Find are bounded by \(x^{2}+y^{2} \leq 2 a x\) and \(y^{2} \geq a x, x \geq 0\).
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