## Trending News

# (數學)微積分的問題!! (積分)

Find the areas of the regions enclosed by the following curves

a. x+y2 = 3 and 4x+y2 = 0 .

b. y=2sinx and y=sin2x, 0<= x <= π

請幫幫我這兩個問題!!

謝謝!!!

### 1 Answer

- Anonymous1 decade agoFavorite Answer
(a) Points of intersection between the curves:

x+y² = 3 and 4x+y² = 0

y² = 3 - x = -4x

3 = -3x

x = -1

y² = 4

y = ±2

Therefore the points of intersection are (-1, 2) and (-1, -2)

So the area can be found as:

A = ∫(-2 → 2) (x_1 - x_2)dy

= ∫(-2 → 2) [(3 - y²) - (-y²/4)] dy

= ∫(-2 → 2) (3 - 3y²/4) dy

= [3y - y³/4] (-2 → 2)

= 8 sq. units

(b) Points of intersection between the curves:

y = 2 sin x and y = sin 2x, 0<= x <= π

2 sin x = sin 2x

2 sin x = 2 sin x cos x

sin x (1 - cos x) = 0

x = 0 or π

So the points of intersection are (0, 0) and (π, 0).

So the area can be found as:

A = ∫(0 → π) (y_1 - y_2)dx

= ∫(0 → π) (2 sin x - sin 2x)dx

= [-2 cos x + 0.5 cos 2x] (0 → π)

= [2 cos x - 0.5 cos 2x] (π → 0)

= 4 sq. units

Source(s): My Maths knowledge- Login to reply the answers