Cho hàm số f(x)={7-4x^2 khi 0<=x<=1 4-x^2 khi x>1 . Diện tích hình phẳng giới

Đáp án B

Xét các phương trình hoành độ giao điểm:

$\begin{array}{l}4-x^2=0\iff \left[\begin{array}{l}x=2\\ x=-2\notin \left(1;+\infty \right)\end{array}\right.\iff x=2;\\ 7-4x^2=0\iff x=\pm \frac{\sqrt{7}}{2}\notin \left[0;1\right]\end{array}$

Suy ra$\begin{array}{l}S=\underset{0}{\overset{1}{\int }}\left|7-4x^2\right|dx+\underset{1}{\overset{2}{\int }}\left|4-x^2\right|dx+\underset{2}{\overset{3}{\int }}\left|4-x^2\right|dx=\underset{0}{\overset{1}{\int }}\left(7-4x^2\right)dx+\underset{1}{\overset{2}{\int }}\left(4-x^2\right)dx+\underset{2}{\overset{3}{\int }}\left(x^2-4\right)dx\\ =\left(7x-\frac{4}{3}{x}^3\right)\left|\begin{array}{l}{}^1\\ {}_0\end{array}\right.+\left(4x-\frac{x^3}{3}\right)\left|\begin{array}{l}{}^2\\ {}_1\end{array}\right.+\left(\frac{x^3}{3}-4x\right)\left|\begin{array}{l}{}^3\\ {}_2\end{array}\right.=7-\frac{4}{3}+\frac{16}{3}-\frac{11}{3}-3+\frac{16}{3}=10.\end{array}$