definite integration

Larry

2011-06-01 3:28 am

圖片參考：http://img202.imageshack.us/img202/8221/65010699.jpg

he figure shows two parabolas withvertices lying on the y-axis. Given that h is the distance between theirvertices, and w is the distance between their points of intersection, show thatthe area bounded by the parabolas is (3/2)hw.

回答 (1)

用戶25975

2011-06-01 4:09 am

✔ 最佳答案

Suppose that the lower parabola has equation y = ax2 + k where k is a constant, then the upper parabola has equation y = bx2 + k + h.

From the given, they intersect at x = -w/2 and w/2 and so:

a(w/2)2 + k = b(w/2)2 + k + h

(a - b)w2/4 = h

a - b = 4h/w2

Thus the area bounded by them is:

∫ (x = -w/2 → w/2) [(bx2 + k + h) - (ax2 + k)] dx

= ∫ (x = -w/2 → w/2) [(b - a)x2 + h] dx

= ∫ (x = -w/2 → w/2) (-4hx2/w2 + h) dx

= [-4hx3/(3w2) + hx] (x = -w/2 → w/2)

= 2[-4hx3/(3w2) + hx] (x = 0 → w/2)

= 2[hw/2 - 4h(w/2)3/(3w2)]

= 2(hw/2 - hw/6) = 2hw/3

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