The area (in sq. units) of the region bounded by the parabola, y=x2+2 and the lines, y=x+1,x=0 and x=3, is:
(A)415
(B)221
(C)417
(D)215✓ Correct
Answer: (D)
Solution
Between x=0 and x=3 the parabola y=x2+2 stays above the line y=x+1, since their difference x2−x+1 has no real roots and is always positive, so the area is ∫03[(x2+2)−(x+1)]dx=∫03(x2−x+1)dx. Evaluating, [3x3−2x2+x]03=9−29+3=215 square units.
JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option