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.