Answer to Question #110584 in Calculus for atta

Question

a curve y=1 - 4/(2x+1)square. the curve intersect the x- axes at A . the normal to the curve at A intersects the y- axes at B.
1) obtain expressions for dy/dx and( upper and lower limit ) |ydx
2) find showing all necessary working find all the area under the curve and normal (full)

Answer 1

$$y=1-{4 \over (2x+1)^2}$$

$$y=0=>1-{4 \over (2x+1)^2}=0=> (2x+1)^2=4$$

$$2x+1=-2\ or \ 2x+1=2$$

$$x=-1.5\ or\ x=0.5$$

Point $A(0.5,0)$

$$y'=(1-{4 \over (2x+1)^2})'=-{4(-2) \over (2x+1)^3}(2x+1)'={16 \over (2x+1)^3}$$

$$y'0.5)={16 \over (2(0.5)+1)^3}=2$$

Fnd the equation of the normal at the point $A(0.5,0)$

$$slope=m=-{1 \over y'(0.5)}=-{1 \over 2}=-0.5$$

$$0=-0.5(0.5)+b=>b=0.25$$

The equation of the normal is $y=-0.5x+0.25.$

Point $B(0,0.25)$

$$S=\displaystyle\int_{0}^{0.5}\big(-0.5x+0.25-(1-{4 \over (2x+1)^2})\big)dx=$$

$$=\big[-0.25x^2-0.75x-{2 \over 2x+1}\big]\begin{matrix} 0.5 \\ 0 \end{matrix}=$$

$$=-0.0625-0.375-1+0+0+2=0.5625(units^2)$$

The area is $0.5625$ square units.