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Area Under the Curves question

2018 · 16 Apr · Shift 1 · Q35
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  5. /2018 · 16 Apr · Shift 1 · Q35

Area Under the Curves question

2018 · 16 Apr · Shift 1 · Q35

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
If the area of the region bounded by the curves, y=x2,y=1xy = {x^2},y = {1 \over x}y=x2,y=x1​ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :
  1. A
    e32{e^{{3 \over 2}}}e23​
  2. B
    43{4 \over 3}34​
  3. C
    32{3 \over 2}23​
  4. D
    e23{e^{{2 \over 3}}}e32​
View written solutionFree

Correct answer: D

  1. Understand the bounded region

    The curves are:

    \qquad y=\frac1x, \qquad y=0, \qquad x=t \; (t>1).$$ First, find where $y=x^2$ and $y=\frac1x$ intersect: $$x^2=\frac1x \implies x^3=1 \implies x=1.$$ At this point, $y=1$. For $x>1$, $$x^2 > \frac1x.$$ Also both curves lie above the $x$-axis. So the region bounded by these curves and lines is made of two parts: - from $x=0$ to $x=1$, between $y=0$ and $y=x^2$; - from $x=1$ to $x=t$, between $y=0$ and $y=\frac1x$.

  2. Form the area expression

    Hence total area is A=∫01x2 dx+∫1t1x dx.A=\int_0^1 x^2\,dx + \int_1^t \frac1x\,dx.A=∫01​x2dx+∫1t​x1​dx.

    Compute each part: ∫01x2 dx=[x33]01=13,\int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1=\frac13,∫01​x2dx=[3x3​]01​=31​, ∫1t1x dx=[ln⁡x]1t=ln⁡t.\int_1^t \frac1x\,dx = [\ln x]_1^t = \ln t.∫1t​x1​dx=[lnx]1t​=lnt.

    Therefore, A=13+ln⁡t.A=\frac13+\ln t.A=31​+lnt.

  3. Use the given condition

    The area is given to be 111 square unit: 13+ln⁡t=1.\frac13+\ln t = 1.31​+lnt=1.

    So, ln⁡t=1−13=23.\ln t = 1-\frac13 = \frac23.lnt=1−31​=32​.

    Exponentiating, t=e2/3.t=e^{2/3}.t=e2/3.

  4. Match with the options

    t=e2/3t=e^{2/3}t=e2/3 which is Option D.

  5. Verification with stored answer

    Stored correct answer: D

    Our derived answer: D

    Hence they agree.

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