1 Marks Question

MCQ 11 Mark

The area of the region bounded by the circle $x^2 + y^2 = 1$ is :

A
$2\pi\text{ sq.}\text{ units}$
✓
$\pi\text{ sq.}\text{ units}$
C
$3\pi\text{ sq.}\text{ units}$
D
$4\pi\text{ sq.}\text{ units}$

Answer

Correct option: B.

$\pi\text{ sq.}\text{ units}$

Question 21 Mark

Sketch the graph of $\text{y}=\sqrt{\text{x}+1}$ in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.

Answer

$\text{y}=\sqrt{\text{x}+1}$ in [0, 4] representa a curve which is part of a parabola x = 4 represrnts a line parallel to y-axis and cutting x-axis at (0, 4)
Enclosed area bound by the cure and lines x = 0 and x = 4 is OABCO consider a vertical strip of length = |y| and width = dx
$\therefore$ Area of appoximating rectangle moves from x = 0 to x = 4
$\Rightarrow \text{A}=\text{Area OABCO}=\int\limits_{0}^{4} |\text{y}|\text{dx}$
$\Rightarrow \text{A}=\int\limits_{0}^{4} \text{y}\text{dx}$
$\Rightarrow \text{A}=\int\limits_{0}^{4}\sqrt{\text{x}+1}\text{dx}$
$\Rightarrow \text{A}=\int\limits_{0}^{4}({\text{x}+1})^\frac{1}{2}\text{dx}$
$\Rightarrow \text{A}=\Bigg[\frac{(\text{x}+1)^\frac{3}{2}}{\frac{3}{2}}\Bigg]^{4}_{0}$
$\Rightarrow \text{A}=\frac{2}{3}\Big(5^\frac{3}{2}-1\Big)\ \text{sq.}\ \text{units}$
$\therefore$ Enclosed area between the cure and given lines $\text{A}=\frac{2}{3}\Big(5^\frac{3}{2}-1\Big)\ \text{sq.}\ \text{units}$

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Question 31 Mark

Using integration, find area of the bounded between the line $x = 2$ and the parabola $y^2 = 8x$.

Answer

$y^2= 8x$ represents a parabola with vertex at origin and axis of symmetry a long the +ve direction of x-axis
x = 2 is line parallel to y-axis.
Let (x, y) be a given point o the parabola, $y^2= 8x$
Since parabola $y^2= 8x$ is symmetric about x-axis,
$\therefore$ Required area = 2(area OCAO)
On slicing the area above x-axis into vertical strios of length = |y| and width = dx
⇒ area of rectangular strip = |y|dx
The approximating rectangle moves between x = 0 and x = 2
So, area A = 2
$\Rightarrow \text{A}=2\int\limits_{0}^{2}\mid\text{y}\mid\text{dx}=2\int\limits_{0}^{2}\text{y}\text{dx}$
$\Rightarrow \text{A}=2\int\limits_{0}^{2}\sqrt{8\text{x}}\text{dx}$
$\Rightarrow \text{A}=2\times2\int\limits_{0}^{2}\sqrt{2\text{x}}\text{dx}=4\sqrt{2}\int\limits_{0}^{2}\sqrt{\text{x}}\text{dx}$
$\Rightarrow \text{A}=4\sqrt{2}\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^{2}_{0}=\frac{8}{3}\sqrt{2}\Big[2^\frac{3}{2}-0\Big]$
$=\frac{8}{3}\times2^{2}=\frac{32}{3}$
$\therefore$ Area $\text{A}=\frac{32}{3}\ \text{sq}.\ \text{units}$

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