1-find-the-area-between-y-2-3x-and-y-x-2-2x-2-0-pi-2-sin-2-cos-2-cos-3-sin-3-2-d-

Question Number 85195 by M±th+et£s last updated on 19/Mar/20

1) f i n d t h e a r e a b e t w e e n

y^{2} = 3 x a n d y = x^{2} - 2 x

2) \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (θ) {c o s}^{2} (θ)}{{({c o s}^{3} (θ) + {s i n}^{3} (θ))}^{2}} d θ

Answered by Rio Michael last updated on 19/Mar/20

at intersection

{(x^{2} - 2 x)}^{2} = 3 x

(x^{4} - 2 (2 x^{3}) + 4 x^{2}) = 3 x

x^{4} - 4 x^{3} + 4 x^{2} - 3 x = 0

\Rightarrow x = 0 or x = 3

A r e a A = \int_{a}^{b} [f (x) - g (x)] d x = \int_{0}^{3} [\sqrt{3} \sqrt{x} - (x^{2} - 2 x)] d x

{= \int_{0}^{3} [\sqrt{3} x^{\frac{1}{2}} - x^{2} + 2 x] = \frac{2 \sqrt{3}}{3} x^{\frac{3}{2}} - \frac{x^{3}}{3} + x^{2}]}_{0}^{3}

A = \frac{2 \sqrt{3}}{3} \sqrt{27} - 9 + 9 = 6 sq units

Commented by M±th+et£s last updated on 19/Mar/20

t h a n k y o u s i r

Answered by mr W last updated on 19/Mar/20

(1)

A r e a = 3 \times 4 - \frac{3 \times 3}{3} - \frac{1 \times 1}{3} - \frac{2 \times 4}{3} = 6

Commented by M±th+et£s last updated on 19/Mar/20

t h a n k y o u s i r