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JEE Main 202430 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2024 · 30 January · Evening Shift · Mathematics · Area Under Curves

The area of the region enclosed by the parabola ( y - 2 ) ^ 2 = x -...

The area of the region enclosed by the parabola (y−2)2=x−1( y - 2{)}^{2} = x - 1, the line x−2y+4=0x - 2y + 4 = 0 and the positive coordinate axes is __________.The area of the region enclosed by the parabola ( y - 2 ) ^ 2 = x - 1 , the line x - 2y + 4 = 0 and the positive

Correct answer

0

Step-by-step solution

Given,

(y−2)2=x−1andx−2y+4=0( y - 2{)}^{2} = x - 1\text{and}x - 2y + 4 = 0

Now, plotting the graph of the above equation, we get,

The area of the region enclosed by the parabola ( y - 2 ) ^ 2 = x - 1 , the line x - 2y + 4 = 0 and the positive

Now, from above diagram we get,

Required area =∫03xdy−= {∫}_{0}^{3}xdy - Area of δABC\delta ABC.
=∫03((y−2)2+1)dy−12×1×2= {∫}_{0}^{3}\left(( y - 2{)}^{2} + 1\right)dy - \dfrac{1}{2} \times 1 \times 2

=∫03(y2−4y+5)dy−1= {∫}_{0}^{3}\left({y}^{2} - 4y + 5\right)dy - 1

=[y33−2y2+5y]03−1= {\left[\dfrac{{y}^{3}}{3} - 2{y}^{2} + 5y\right]}_{0}^{3} - 1

=9−18+15−1=5= 9 - 18 + 15 - 1 = 5

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