Search any question & find its solution
Question:
Answered & Verified by Expert
Area lying between the curves $\mathrm{y}^2=\mathbf{4 x}$ and $\mathrm{y}=\mathbf{2 x}$.
(a) $\frac{2}{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{4}$
(d) $\frac{3}{4}$
(a) $\frac{2}{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{4}$
(d) $\frac{3}{4}$
Solution:
1002 Upvotes
Verified Answer
(b) The curve is $y^2=4 x$ and the line is $y=2 x$
from (i) and (ii): $\left(4 x^2\right)=4 x \Rightarrow x=0$, or $x=1$
curves (i) and (ii)
intersect at $\mathrm{O}(0,0), \mathrm{A}(1,2)$
Area of region $\mathrm{OACO}$
$=$ Area of region $\mathrm{OMACO}$
- area of OMA
Now Area of region OMACO
$=$ Area of region bounded by $\mathrm{OCA}$ $=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{d} x=2 \cdot \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_0^1=\frac{4}{3} \times 1=\frac{4}{3}$ sq. units.
Area of $\triangle \mathrm{OMA}=$ Area of region bounded by OA: $\mathrm{y}=2 \mathrm{x}$, $x=1$ and $x$-axis. $=\int_0^1 2 x d x=\left[x^2\right]_0^1=1$
Putting these values in (iii), we get :
$\therefore$ Area of region $\mathrm{OMACO}=\frac{4}{3}-1=\frac{1}{3}$ sq units.
from (i) and (ii): $\left(4 x^2\right)=4 x \Rightarrow x=0$, or $x=1$
curves (i) and (ii)
intersect at $\mathrm{O}(0,0), \mathrm{A}(1,2)$
Area of region $\mathrm{OACO}$
$=$ Area of region $\mathrm{OMACO}$
- area of OMA
Now Area of region OMACO
$=$ Area of region bounded by $\mathrm{OCA}$ $=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{d} x=2 \cdot \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_0^1=\frac{4}{3} \times 1=\frac{4}{3}$ sq. units.
Area of $\triangle \mathrm{OMA}=$ Area of region bounded by OA: $\mathrm{y}=2 \mathrm{x}$, $x=1$ and $x$-axis. $=\int_0^1 2 x d x=\left[x^2\right]_0^1=1$
Putting these values in (iii), we get :
$\therefore$ Area of region $\mathrm{OMACO}=\frac{4}{3}-1=\frac{1}{3}$ sq units.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.