We have,
y2=8x
y=x
Point of intersection of curve and a line is A(8,8).

Required area
α=∫28(8x−x)dx
⇒α=[342x23−2x2]28
⇒α=[(342(22)3−264)−(342(2)3−24)]
⇒α=3112−30
⇒α=322
So, 3α=22
JEE Main 2023 — Mathematics Calculus
Let α be the area of the larger region bounded by the curve y2=8x and the lines y=x and x=2, which lies in the first quadrant. Then the value of 3α is equal to
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