a) Diện tích hình phẳng giới hạn bởi \(y=x;y=2x;x=0;x=2\)
\(S=\int\limits^2_0\left(2x-x^2\right)dx\left(2x>x^2;x\in\left[0;2\right]\right)\)
\(\Rightarrow S=\left[x^2-\dfrac{x^3}{3}\right]^2_0\)\(=4-\dfrac{8}{3}=\dfrac{4}{3}\left(đvdt\right)\)
b) Diện tích hình phẳng giới hạn bởi \(y=x^2-x+3;y=2x+1;x=0;x=2\)
\(S=\int\limits^1_0\left[\left(x^2-x+3\right)-\left(2x+1\right)\right]dx+\int\limits^2_1\left[-\left(x^2-x+3\right)+\left(2x+1\right)\right]dx\)
\(\Rightarrow S=\int\limits^1_0\left(x^2-3x+2\right)dx+\int\limits^2_1\left(-x^2+3x-2\right)dx\)
\(\Rightarrow S=\left[\dfrac{x^3}{3}-\dfrac{3x^2}{2}+2x\right]^1_0+\left[-\dfrac{x^3}{3}+\dfrac{3x^2}{2}-2x\right]^2_1\)
\(\Rightarrow S=\left(\dfrac{1}{3}-\dfrac{3}{2}+2\right)-0+\left(-\dfrac{8}{3}+6-4\right)-\left(-\dfrac{1}{3}+\dfrac{3}{2}-2\right)=\dfrac{5}{6}-\dfrac{4}{6}+\dfrac{5}{6}=1\left(đvdt\right)\)
