Advanced Math/Pure maths

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Expert: - 10/27/2010


Question

graph
The figure shows the curve with the equation y=-2x^2+6X-4. The curve meets the y-axis at the point A and the x-axis at point B and C.
Question--> Show that the area of the shaded region bounded by the curve, the y- axis and the line ED is 0.984375 square units.

Answer
To find C we want the zeros of y=-2x^2+6X-4.
Solving 0 = -2x^2+6X-4 , we get x=1 or x=2  , so C=(2,0)

To find the normal line to the parabola at C = (2,0) , we need the slope of the tangent at x = 2.

y' = -4x + 6

When x=2 , y' = -2

The normal line is perpendicular to the tangent so its slope is
-(1/-2) = 1/2  

The line EC has slope 1/2 and passes through (2,0)

An equation for EC is then (y-0)/(x-2) = 1/2  or y = (1/2)x - 1

To find D we get the intersection of the line y = (1/2)x - 1 with the parabola y = -2x^2+6x-4


(1/2)x - 1 = -2x^2+6x-4


Solving this gives x = 2 or x = 3/4

So the x coordinate of D is 3/4


We want the area between the curves y = (1/2)x - 1 and
y = -2x^2 + 6x - 4 when x is between 0 and 3/4

The area is

S [(1/2)x - 1] - [-2x^2 + 6x - 4] dx  with limits of integration 0
and 3/4

S 2x^2 - (11/2)x + 3 dx   with limits of integration 0
and 3/4


an anti derivative for 2x^2 - (11/2)x + 3 is

(2/3)x^3 - (11/4)x^2 + 3x


Since this is 0 when x = 0 , we need only evaluate this at x = 3/4

(2/3)(3/4)^3 - (11/4)(3/4)^2 + (3)(3/4) = 0.984375 , as you can easily check  

Socrates

Expertise

I can answer any questions from the standard four semester Calulus sequence. Derivatives, partial derivatives, chain rule, single and multiple integrals, change of variable, sequences and series, vector integration (Green`s Theorem, Stokes, and Gauss) and applications. Pre-Calculus, Linear Algebra and Finite Math questions are also welcome.

Experience

Ph.D. in Mathematics and many years teaching undergraduate courses at three state universities.

Education/Credentials
B.S. , M.S. , Ph.D.

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