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Application of Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Integrals1 Mark
MCQ
Area lying between the curves $y^2=4 x$ and $y=2 x$ is :
  • A
    $\frac{2}{3}$
  • ✓
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{3}{4}$
✓

Answer

Correct option: B.
$\frac{1}{3}$
The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations,
$y^2=4 x $ and $y=2 x$
$\Rightarrow(2 x)^2=4 x$
$\Rightarrow 4 x^2=4 x$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0$ or $x=1$
$\Rightarrow y=0$ or $y=2$
Thus, $O(0, 0)$ and $A(1, 2)$ are the points of intersection of the parabola and straight line shaded area is the required area.
Using the horizontal strip method, shaded area
$= \int\limits^2_0|\text{x}_2-\text{x}_1|\text{dy}$
$=\int\limits^2_0\Big[\Big(\frac{\text{y}}{2}\Big)-\Big(\frac{\text{y}^2}{4}\Big)\Big]\text{dy}$
$=\Big[\frac{1}{2}\Big(\frac{\text{y}^2}{2}\Big)-\frac{1}{4}\Big(\frac{\text{y}^3}{3}\Big)\Big]^2_0$
$=\frac{1}{4}(2)^2-\frac{1}{12}(2^3)-0$
$= 1 -\frac{8}{12}$
$= \frac{12-8}{12}$
$=\frac{1}{3}\text{ sq. units}$

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