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Area Under the Curves question

2005 · Shift 0 · Q80
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Area Under the Curves question

2005 · Shift 0 · Q80

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area enclosed between the curve y=log⁡e(x+e)y = {\log _e}\left( {x + e} \right)y=loge​(x+e) and the coordinate axes is :
  1. A
    111
  2. B
    222
  3. C
    333
  4. D
    444
View written solutionFree

Correct answer: A

  1. Find where the curve meets the axes

The curve is y=ln⁡(x+e).y=\ln(x+e).y=ln(x+e).

  • Intersection with the yyy-axis: put x=0x=0x=0 y=ln⁡(e)=1.y=\ln(e)=1.y=ln(e)=1. So the point is (0,1)(0,1)(0,1).

  • Intersection with the xxx-axis: put y=0y=0y=0 ln⁡(x+e)=0\ln(x+e)=0ln(x+e)=0 x+e=1x+e=1x+e=1 x=1−e.x=1-e.x=1−e. So the point is (1−e,0)(1-e,0)(1−e,0).

Thus, the region enclosed by the curve and the coordinate axes lies between x=1−ex=1-ex=1−e and x=0x=0x=0, above the xxx-axis.

  1. Set up the area integral

Required area is A=∫1−e0ln⁡(x+e) dx.A=\int_{1-e}^{0} \ln(x+e)\,dx.A=∫1−e0​ln(x+e)dx.

Let u=x+e⇒du=dx.u=x+e \quad \Rightarrow \quad du=dx.u=x+e⇒du=dx.

When x=1−ex=1-ex=1−e, u=1u=1u=1. When x=0x=0x=0, u=eu=eu=e.

So, A=∫1eln⁡u du.A=\int_{1}^{e} \ln u\,du.A=∫1e​lnudu.

  1. Evaluate the integral

We know ∫ln⁡u du=uln⁡u−u.\int \ln u\,du=u\ln u-u.∫lnudu=ulnu−u.

Hence, A=[uln⁡u−u]1e.A=\left[u\ln u-u\right]_{1}^{e}.A=[ulnu−u]1e​.

Now,

  • At u=eu=eu=e: eln⁡e−e=e−e=0.e\ln e-e=e-e=0.elne−e=e−e=0.
  • At u=1u=1u=1: 1ln⁡1−1=0−1=−1.1\ln 1-1=0-1=-1.1ln1−1=0−1=−1.

Therefore, A=0−(−1)=1.A=0-(-1)=1.A=0−(−1)=1.

  1. Check options

The enclosed area is 1.\boxed{1}.1​.

So the correct option is A.

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