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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Area Under Curves

The area (in sq. units) in the first quadrant bounded by the...

The area (in sq. units) in the first quadrant bounded by the parabola, y=x2+1y=x^{2}+1, the tangent to it at the point (2,5) and the coordinate axes is :The area (in sq. units) in the first quadrant bounded by the parabola, y=x^ 2 +1 , the tangent to it at the point (2,5)

Options

  1. A83\frac{8}{3}
  2. B3724\frac{37}{24}
  3. C18724\frac{187}{24}
  4. D143\frac{14}{3}

Correct answer

B. 3724\frac{37}{24}

Step-by-step solution

The area (in sq. units) in the first quadrant bounded by the parabola, y=x^ 2 +1 , the tangent to it at the point (2,5)
The equation of parabola x2=y−1x^{2}=y-1 The equation of tangent at (2,5) to parabola is y−5=y-5=\left(\frac{d y}{d x}\right)_{(2,5)}(x-2) y-5=4(x-2) 4 x-y=3Then,therequiredareaThen, the required area=\int_{0}^{2}\left\{\left(x^{2}+1\right)-(4 x-3)\right\} d x-AreaofArea of\Delta A O D =\int_{0}^{2}\left(x^{2}-4 x+4\right) d x-\frac{1}{2} \times \frac{3}{4} \times 3 =\left[\frac{(x-2)^{3}}{3}\right]_{0}^{2}-\frac{9}{8}=\frac{37}{24}
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