Question

Let y=f(x) be the given curve and $x=a$, $x=b$ be two ordinates then area bounded by the curve $y=f\left(x\right)$, the axis of x between the ordinates $x=a$ & $x=b$, is given by definite integral
${\int }_a^{b}ydx$ or ${\int }_a^{b}f\left(x\right)dx$ and the area bounded by the curve $x=f\left(y\right)$, the axis of y & two abscissae $y=c$ & $y=d$ is given by ${\int }_c^{d}xdy$ or ${\int }_c^{d}f\left(x\right)dy$. Again if we consider two curves $y=f\left(x\right)$, $y=g\left(x\right)$ where $f\left(x\right)\ge g\left(x\right)$ in the interval [a, b] where $x=a$ & $x=b$ are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by
${\int }_a^b[f\left(x\right)-g\left(x\right)]dx$
On the basis of above information answer the following questions.

The area bounded by parabolas $y=x^2+2x+1$ & $y=x^2-2x+1$ and the line $y=\frac{1}{4}$ is equal to

• A. $\frac{2}{3}$ square unit
• B. $\frac{1}{3}$ square unit
• C. $\frac{3}{2}$ square unit
• D. $\frac{1}{2}$ square unit

Answer 1

Answer: (B)

$\frac{1}{3}$ square unit

Given curve is $y=x^2+2x+1={(x+1)}^2$ (i)
is upward parabola with vertex at (-1, 0) meet y-axis at (0, 1) and the curve $y=x^2-2x+1={(x-1)}^2$ (ii)
is also upward parabola with vertex at (1, 0) meet, y-axis at (0, 1)
Also $y=\frac{1}{2}$ (iii)
A line parallel to x-axis meeting(i) at $(-\frac{1}{2},\frac{1}{4})$, $(-\frac{3}{2},\frac{1}{4})$ & meeting (ii) at $(\frac{3}{2},\frac{1}{4})$, $(\frac{1}{2},\frac{1}{4})$
Required area is the shaded region given by 2 Area (LTNL) (by symmetry)
$=2{\int }_0^{1/2}\{{(x-1)}^2-\frac{1}{4}\}dx=2{[\frac{{(x-1)}^3}{3}-\frac{x}{4}]}_0^{1/2}$
$=2[(-\frac{1}{24}-\frac{1}{8})+\left(\frac{1}{3}\right)]=2[-\frac{1}{6}+\frac{1}{3}]=\frac{1}{3}$ square unit