tìm S : $y=\sqrt{x} ,và  x+y-2=0, y=0$
tìm S : $y=x^{2} , y=4x^{2} , y=4
$

b, Đặt $(C_1): y=x^2, \quad(C_2): y=4x^2, \quad(C_3): y=4, \quad$
Ta có:
$(C_1) \cap (C_2) : x^2=4x^2 \Leftrightarrow x=0$
$(C_3) \cap (C_2) : 4=4x^2 \Leftrightarrow x=\pm 1$
$(C_1) \cap (C_3) : x^2=4 \Leftrightarrow x=\pm2$
Để minh họa rõ ràng bạn nên vẽ đồ thị của ba hàm số này. Ta có thể thấy rằng theo tính chất đối xứng thì diện tích cần tìm bằng hai lần diện tích nằm ở góc phần tư thứ nhất.
$S=2\int\limits_{0}^{1}\left| { x^2-4x^2} \right|dx+2\int\limits_{1}^2\left| { 4-x^2} \right|dx$
    $=2\int\limits_{0}^{1}3x^2dx+2\int\limits_{1}^2(4-x^2)dx$
    $=2 +\dfrac{10}{3} =\dfrac{16}{3} $
a) Viết lại $\begin{cases}x=y^2 \\ x=-y+2 \end{cases}$
Giải PT $ y^2=2-y\Leftrightarrow y^2+y-2=0\Leftrightarrow  y=1$, do $y=\sqrt x \ge 0.$
Kết hợp với $y=0$ ta có
$S=\int\limits_{0}^{1}\left| {(y^2)-(-y+2)} \right|dy=\int\limits_{0}^{1}\left| {y^2+y-2} \right|dy=-\int\limits_{0}^{1}(y^2+y-2)dy=\dfrac{7}{6}$

Bạn cần đăng nhập để có thể gửi đáp án

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