Finding Area of Shaded Region (Calculus)?

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You have shown us no shaded region, but I suppose you mean the region enclosed by the given curves.

x = 4 - y²
x = y - 2

That is a parabola and a line. If they intersect, the enclosed region is a parabola segment.

4 - y² = y - 2
y² + y - 6 = 0
y = -3 or y = 2

The points of intersection: A(-5, -3), B(0, 2)
Midpoint of the chord: M(-5/2, -1/2)
Central diameter: y = -1/2

Find point C, where the central diameter meets the parabola.

y = -1/2
x = 4 - (-1/2)² = 15/4
C(15/4, -1/2)

area(∆ACB) = 1/2[15/4 - (-5/2)][2 - (-3)] = 1/2(25/4)(5) = 125/8

area(segment ACB) = 4/3(125/8) = 125/6...Show more

how can we tell what the shaded area is ??

but I can guess...

A = ∫(4–y²) dy –  ∫ (y-2) dy   between y = –3 and y = +2

A = (4y – (1/3)y³) – ((1/2)y² – 2y) between y = –3 and y = +2
A = (4•2 – (1/3)2³) – ((1/2)2² – 2•2) – (4(–3) – (1/3)(–3)³) + ((1/2)(–3)² – 2(–3))
A = (8 – (8/3)) – (2 – 4) – ((–12) + (27/3)) + ((9/2) + 6)
A = (8 – (8/3)) – (– 2) – ((–12) + 9) + ((9/2) + 6)
A = (24/3 – 8/3) + 2 – (–3) + (9/2 + 12/2)
A = (16/3) + 2 + 3 + (21/2)
A = (16/3) + 5 + (21/2)
A = (32/6) + (30/6) + (63/6)
A = 125/6 = 20.833..

but check the arith......Show more