If the line x = α divides the area of the region R = { x , y ∈ R 2 : x 3 ≤ y ≤ x , 0 ≤ x ≤ 1 } into two equal parts, then

A = ∫ 0 1 x - x 3 d x = 1 4 Area of curvilinear triangle O P Q = A 2 = 1 8 ⇒ ∫ 0 1 x - x 3 d x = 1 8 ⇒ 2 α 4 - 4 α 2 + 1 = 0 ⇒ α 2 - 1 2 = 1 2 ⇒ α 2 = 2 - 1 2 so α lies between 0.5 and 1

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2017

MathematicsMediummultiple choice

If the line $x = α$ divides the area of the region $R = {(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1}$ into two equal parts, then

Options:(select one or more)

Answer:

A, D

Solution:

$A = \int_{0}^{1} (x - x^{3}) d x = \frac{1}{4}$

Area of curvilinear triangle $O P Q = \frac{A}{2} = \frac{1}{8}$

$\Rightarrow \int_{0}^{1} (x - x^{3}) d x = \frac{1}{8} \Rightarrow 2 α^{4} - 4 α^{2} + 1 = 0$

$\Rightarrow {(α^{2} - 1)}^{2} = \frac{1}{2} \Rightarrow α^{2} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

so α lies between 0.5 and 1

Stream:JEE_ADVSubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2017

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